Make r the subject of x=e+r/d

Answer:

4.2 miles

Step-by-step explanation:

Martina runs 6 miles in 50 minutes, we have to find the rate at which she is running. Put it in a fraction [tex]\frac{6}{50}[/tex] or she runs 6 miles every 50 minutes. When we divide we get that she is running, 0.12 miles evrey minute. The question askes us how far Matina will run in 35 minutes, so we multiply 0.12 by 35, and we get that she will run 4.2 miles.

Answer:

#1 is the answere

Step-by-step explanation:

Check the sides

Answer:

40

Step-by-step explanation:

Answer:

24-10-10=4

4/2=2

10*2=20

Step-by-step explanation:

Answer: $3.60 per case

Step-by-step explanation: $21.60 divided by 6 = 3.60

The equivalence of the expressions n × ([tex]{}^{(n-1)}C_2[/tex]) and [tex]{}^nC_2[/tex] × (n − 2) has been proven mathematically.

To prove the equivalence of the expressions n × ([tex]{}^{(n-1)}C_2[/tex]) and [tex]^{n}C_2[/tex] × (n − 2), we can demonstrate that they yield the same result.

First, let's simplify each expression:

n × ([tex]{}^{(n-1)}C_2[/tex]) = n × [(n-1)! / 2!(n-1-2)!]

= n × [(n-1)! / 2!(n-3)!]

= n × [(n-1)(n-2) / 2]

= n × (n² - 3n + 2) / 2

= (n³ - 3n² + 2n) / 2

[tex]{}^nC_2[/tex] × (n − 2) = [n! / 2!(n-2)!] × (n-2)

= [n! / 2!(n-2)!] × (n-2)

= [(n)(n-1)(n-2)! / 2!(n-2)!] × (n-2)

= [(n)(n-1)] / 2

= (n² - n) / 2

By comparing the two simplified expressions, we can see that (n³ - 3n² + 2n) / 2 is equal to (n² - n) / 2.

Hence, we have proven that n × ([tex]{}^{(n-1)}C_2[/tex]) is equivalent to [tex]{}^nC_2[/tex] × (n − 2).

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Answer:

Werid I don’t know

Step-by-step explanation:

Answer:

x=5/6 and y=8/5

(5/6 , 8/5)

Step-by-step explanation:

You can use the elimination method! The point of this method is to add or subtract one equation from the other, multiplying them by constants if necessary, to cancel out one variable so you can solve for the other. Then, you can plug the value you got for your solution into either equation to solve for the other. I'll demonstrate.

Look at the two equations and the coefficients of both variables. You have 12x and -6x -- 6*2=12, so this is perfect. (You could also eliminate the y because 5*3=15, but I'll show you by eliminating the x instead.)

Here's what that looks like:

(-6x+1y=3)2

-12x+10y=6

So we just multiplied the second equation by 2 on both sides. Let's see how that helps us.

12x+15y=34

-12x+10y=6

If we add the two equations now, x will be canceled out and we can solve for y.

   12x+15y=34

+(-12x+10y=6)

___________

     0x+25y=40

           25y=40

0x=0, so we can get rid of the x. Now, we need to solve for y.

25y=40

y=40/25

You probably know how to simplify fractions, so divide both the numerator and the denominator by 5 to get y=8/5.

Now you can use this value in either equation and solve for x. I'll use the first. (This is called substitution.)

12x+15(8/5)=34

12x+3(8)=34 <-- What I did here is cancel out the 5 in the denominator with 15 to leave 3, because 5*3=15.

12x+24=34

12x=10

x=10/12

x=5/6 (Divide the numerator and denominator by 2.)

You can write your answer as a point, too. (5/6 , 8/5)

Answer:

10/d

Step-by-step explanation:

chfhnclfsiojjcllkn

Answer:

Step-by-step explanation:

Professional dancers cause I’ve danced since I was 3 and I’ve always wanted to do it. And I wanna model on the side

Answer:

Detective/Forensics scientist

Step-by-step explanation:

I want to be a detective/ forensic scientist because I am interested in science and investigating.

Answer:

no don't ever click on links okay.

Step-by-step explanation:

This paper will discuss the mathematics involved in determining the least mark that Jethro needs to get for the last test in order to end up with a mean mark of 77 over all 6 tests.

Jethro has sat 5 tests and each test was marked out of 100. Therefore, Jethro’s marks for the 5 tests can be represented by x1, x2, x3, x4, and x5. Jethro’s mean mark for the 5 tests is 74, which can be represented by the mathematical equation: (x1 + x2 + x3 + x4 + x5) / 5 = 74.

Jethro needs to sit one more test that is also marked out of 100 and he wants his mean mark for all 6 tests to be at least 77. Therefore, the least mark that Jethro needs to get for the last test can be found by rearranging the equation to: (x1 + x2 + x3 + x4 + x5 + x6) / 6 = 77. This can be written in its simplified form as: 6x6 = 77(x1 + x2 + x3 + x4 + x5).

In order to find the least mark that Jethro needs to get for the last test, x6, the other terms must be known. Since Jethro has already sat the 5 tests the marks for these tests are known, so they can be added together. This results in: x6 = 77(x1 + x2 + x3 + x4 + x5) / 6. Therefore, Jethro needs to get a mark of at least 77.6 (rounded to the nearest tenth) on the last test in order to end up with a mean mark of 77.

Answer:

True

False

False

Step-by-step explanation:

The α = 0.01 level of significance, we can conclude that the average alcohol consumption of college students is significantly higher than that of the average American population.

To determine if the average alcohol consumption of college students is significantly different from the average consumption of the average American, we can conduct a hypothesis test.

Let's set up the hypotheses:

Null hypothesis (H0): The average alcohol consumption of college students is equal to the average American consumption. (μ = 81)

Alternative hypothesis (H1): The average alcohol consumption of college students is greater than the average American consumption. (μ > 81)

We can use a one-sample t-test to analyze the data. Since we don't have information about the population standard deviation, we'll use the t-distribution and the sample standard deviation instead.

Given that the sample mean (x) of the 10 randomly selected college students is 97.7 liters and the sample standard deviation (s) is 23 liters, we can calculate the t-statistic using the following formula:

t = (x - μ) / (s / √n)

Where:

x = sample mean

μ = population mean

s = sample standard deviation

n = sample size

Plugging in the values, we get:

t = (97.7 - 81) / (23 / √10)

Calculating this expression gives us the t-value.

However, we also need to determine the critical value for the test based on the significance level (α = 0.01) and the degrees of freedom = n - 1.

Since we have 10 randomly selected college students = 10 - 1 = 9.

To find the critical value, we can consult the t-distribution table. With α = 0.01 and df = 9, the critical t-value is approximately 2.821.

Comparing the calculated t-value to the critical t-value, we can draw a conclusion. If the calculated t-value is greater than the critical t-value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

Now let's calculate the t-value:

t = (97.7 - 81) / (23 / √10)

≈ 16.700

Since the calculated t-value (16.700) is much greater than the critical t-value (2.821), we can reject the null hypothesis.

Therefore, based on the given data and the α = 0.01 level of significance, we can conclude that the average alcohol consumption of college students is significantly higher than that of the average American population.

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Answer:

the time taken for the radioactive element to decay to 1 g is 304.8 s.

Step-by-step explanation:

Given;

half-life of the given Dubnium = 34 s

initial mass of the given Dubnium, m₀ = 500 grams

final mass of the element, mf = 1 g

The time taken for the radioactive element to decay to its final mass is calculated as follows;

[tex]1 = 500 (0.5)^{\frac{t}{34}} \\\\\frac{1}{500} = (0.5)^{\frac{t}{34}}\\\\log(\frac{1}{500}) = log [(0.5)^{\frac{t}{34}}]\\\\log(\frac{1}{500}) = \frac{t}{34} log(0.5)\\\\-2.699 = \frac{t}{34} (-0.301)\\\\t = \frac{2.699 \times 34}{0.301} \\\\t = 304.8 \ s[/tex]

Therefore, the time taken for the radioactive element to decay to 1 g is 304.8 s.

The time required to decay 500 grams to 1 gram is 304.8 seconds and this can be determined by using the given data.

Given :

Dubnium-262 has a half-life of 34 s.

Final mass = 1 gram

Initial mass = 500 gram

Time taken by a radioactive element to decay is:

[tex]1 = 500(0.5)^{\frac{t}{34}}[/tex]

Simplify the above equation.

[tex]\rm \dfrac{1}{500} = (0.5)^{\frac{t }{34}}[/tex]

Now, take the log on both sides in the above equation.

[tex]\rm log(0.002 ) = \dfrac{t}{34}\times log(0.5)[/tex]

[tex]\rm \dfrac{log(0.002)}{log(0.5)} \times 34 = t[/tex]

t = 304.8 sec

So, the time required to decay 500 grams to 1 gram is 304.8 seconds.

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Answer:

Step-by-step explanation:

a) 9.80- 98/10

b) 7.3- 73/10

a) 7/100- 0.07

b) 82/10- 8.2

Hope this helps

Fraction:

a) 98/100

b) 73/10

Decimal:

a) 0.07

b) 8.2

The mean of the given probability distribution is 2.4.

To find the mean of a probability distribution, we multiply each value of x by its corresponding probability and then sum them up. Using the provided data:

x: 0, 1, 2, 3, 4

P(x): 0.0017, 0.3421, 0.065, 0.4106, 0.1806

mean = 0(0.0017) + 1(0.3421) + 2(0.065) + 3(0.4106) + 4(0.1806)

     = 0 + 0.3421 + 0.13 + 1.2318 + 0.7224

     = 2.4263

Therefore, the mean of the given probability distribution is approximately 2.4 (rounded to one decimal place).

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Answer: You will have one million dollars in the account after approximately 84.89 years.

APR is a yearly percentage rate that reflects the actual cost of borrowing on loans and investments. The APR is the rate of interest that must be charged on the balance of a savings account to attain a certain goal in the specified time period. The formula for compound interest is used in this case. The formula for compound interest is:A=P(1+r/n)^(nt)Where: A = amount P = principal (initial amount) r = annual interest rate (as a decimal) n = number of times interest is compounded per year t = number of years In this scenario: A = $1,000,000P = $188.00r = 0.028n = 1 (compounded once per year)t = unknown. Now let's solve for t:1,000,000 = 188(1 + 0.028/1)^(1t)ln (5,319.15) = t ln (1.028) ln (5,319.15) = 0.028t84.89 years = t Therefore, it will take approximately 84.89 years to reach one million dollars in the account.

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The estimated value of "a" using the method of moments is 48.00.

The method of moments is a technique used to estimate the parameters of a probability distribution by equating the sample moments to their theoretical counterparts. In this case, we'll equate the sample mean to the theoretical mean of the uniform distribution.

The theoretical mean of a uniform distribution with density function f(x) = 1/a is given by (a + 0) / 2 = a / 2.

To estimate "a," we'll equate the sample mean to a / 2 and solve for "a":

Sample mean = (41 + 37 + 48 + 32 + 43 + 21 + 29 + 22 + 40 + 28 + 22 + 29 + 38 + 23 + 24) / 15

           = 34.13 (rounded to two decimal places)

Setting this equal to a / 2, we have:

34.13 = a / 2

Solving for "a," we multiply both sides by 2:

a = 2 * 34.13

 ≈ 68.26

Rounding "a" to two decimal places:

a ≈ 68.26 ≈ 68.00

Using the method of moments, the estimated value of "a" is approximately 68.00. This suggests that the data points were sampled from a uniform distribution with a maximum value of around 68.

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The Laplace transform of f(t) = t - cos(3t) + [tex]e^{7t}[/tex]+ (t - 1)² is:

L{f(t)} = (1 - 2/s + 1/s²) - (s/(s² + 9)) + 1/(s - 7) + 2/s³

To find the inverse Laplace transform of the expression 232 + 7s + 14 (32 + 2x + 10) (5 + 1), we need to break it down into simpler terms and apply the inverse Laplace transform individually.

Given expression: 232 + 7s + 14 (32 + 2x + 10) (5 + 1)

Let's simplify the expression first:

232 + 7s + 14 (32 + 2x + 10) (5 + 1) = 232 + 7s + 14 ×42× 6

Simplifying further:

232 + 7s + 3528

Now we have a simple expression. To find the inverse Laplace transform, we can use the linearity property of the Laplace transform.

Inverse Laplace transform of 232 is 232 × δ(t), where δ(t) is the Dirac delta function.

Inverse Laplace transform of 7s is 7 δ'(t), where δ'(t) is the derivative of the Dirac delta function.

The inverse Laplace transform of a constant times the Dirac delta function is given by multiplying the constant with the shifted unit step function.

Inverse Laplace transform of 14 ×3528 is 14× 3528 × u(t), where u(t) is the unit step function.

Therefore, the inverse Laplace transform of the given expression is:

Inverse Laplace transform of (232 + 7s + 14 (32 + 2x + 10) (5 + 1)) = 232 ×δ(t) + 7 δ'(t) + 14×3528× u(t)

To find the Laplace transform of f(t) = t - cos(3t) + [tex]e^{7t}[/tex] + (t - 1)², we will apply the properties of Laplace transforms to each term individually.

Laplace transform of t:

The Laplace transform of t, denoted as L{t}, is given by 1/s^2.

Laplace transform of cos(3t):

The Laplace transform of cos(3t), denoted as L{cos(3t)}, is given by s/(s²+ 9).

Laplace transform of [tex]e^{7t}[/tex]:

The Laplace transform of [tex]e^{7t}[/tex], denoted as L{[tex]e^{7t}[/tex]}, is given by 1/(s - 7).

Laplace transform of (t - 1)²:

We can expand (t - 1)²to t² - 2t + 1 and then apply the linearity property of Laplace transforms.

Laplace transform of t²:

The Laplace transform of t², denoted as L{t²}, is given by 2/s³.

Laplace transform of 2t:

The Laplace transform of 2t, denoted as L{2t}, is given by 2/s².

Laplace transform of 1:

The Laplace transform of 1, denoted as L{1}, is given by 1/s.

Using the linearity property of Laplace transforms, we can add the transforms of each term.

Laplace transform of f(t):

L{t} - L{cos(3t)} + L{[tex]e^{7t}[/tex]} + L{(t - 1)²}

= 1/s² - s/(s² + 9) + 1/(s - 7) + 2/s³ - 2/s² + 1/s

= (1 - 2/s + 1/s²) - (s/(s² + 9)) + 1/(s - 7) + 2/s³

Therefore, the Laplace transform of f(t) = t - cos(3t) + [tex]e^{7t}[/tex]+ (t - 1)² is:

L{f(t)} = (1 - 2/s + 1/s²) - (s/(s² + 9)) + 1/(s - 7) + 2/s³

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The probability that a randomly selected student scores more than 82% on the physics test, we can use the standard normal distribution and the given mean and standard deviation.

The z-score formula is given by z = (x - μ) / σ, where z represents the z-score, x is the observed value, μ is the mean, and σ is the standard deviation. In this case, the observed value is 82%, the mean is 75%, and the standard deviation is 6.5%. Plugging these values into the formula, we calculate the z-score as z = (0.82 - 0.75) / 0.065 = 1.0769.

Next, we need to find the area to the left of the z-score in the standard normal distribution table or using a calculator. The area to the left of 1.0769 corresponds to the probability of scoring less than 82%. Let's assume this area is P(z < 1.0769).

The probability of scoring more than 82%, we subtract P(z < 1.0769) from 1: P(z > 1.0769) = 1 - P(z < 1.0769).

Using a standard normal table or a calculator, we can find P(z < 1.0769) to determine the probability.

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Answer:

[tex]x^{\frac{5}{3} }[/tex]

Step-by-step explanation:

[tex]\frac{2}{3} *\frac{5}{2} \\\\\frac{5}{3}[/tex]

Answer:

They are opposite angles. If you rotate the figure SQM 180 degrees counter clockwise, they will be the exact same triangle and therefore the exact same measure.

Step-by-step explanation:

it's 40.00 beacuase justo subtract

9514 1404 393

Answer:

  Chris's gym charges more

Step-by-step explanation:

The difference between 2 rentals and 1 rental at Chris's gym is ...

  $55.50 -52.75 = $2.75 . . . . cost of court rental at Chris's gym

This $2.75 cost at Chris's gym is higher than the corresponding $2.00 cost at Tyrell's gym.

Chris's gym charges more to reserve the basketball courts.

The probability that it will not snow on January 1st in that town in a given year is 0.815.

Based on the meteorological records, A probability forecast includes a numerical expression of uncertainty about the quantity or event being forecast. Ideally, all elements (temperature, wind, precipitation, etc.)

The probability that it will not snow in a certain town on January 1st in a given year is 0.815. Here's how to arrive at the answer:Given that the probability of snowing on January 1st in that town is 0.185. Then, the probability of not snowing on January 1st is 1 - 0.185 = 0.815.

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The given information probability of it not snowing on January 1st of a given year in that town is 0.815.

Therefore, the probability of it not snowing on January 1st of a given year in that town is 0.815.

Here's how to solve the problem: Given: The probability of it snowing on January 1st of a given year in that town is 0.185. The complement of the probability of it snowing on January 1st is the probability of it not snowing on January 1st of a given year in that town, which is:

P(not snowing on January 1st) = 1 - P(snowing on January 1st)

P(not snowing on January 1st) = 1 - 0.185

P(not snowing on January 1st) = 0.815

Therefore, the probability of it not snowing on January 1st of a given year in that town is 0.815.

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Answer:

No, the ribbon would not be long enough to go around the edge of the cake.

Circumference ≈ 81.7 cm

Step-by-step explanation:

1) Calculate the circumference of the cake

Circumference - the length around a circle

[tex]C=2\pi r[/tex] where r is the radius

Plug 13 in as the radius

[tex]C=2\pi (13)\\C=26\pi\\C= 81.7[/tex]

Therefore, the circumference of the cake is approximately 81.7 cm.

Because the ribbon is only 36 cm long, it would not be enough to go around the edge of the cake.

I hope this helps!

Answer:

Stefan spent 108.9$ and made 12.1$

Step-by-step explanation:

Hey I'm Aiden I can help you, I will take Brainlest, Your welcome :)

take 10% of 121 and subtract it from 121 and you get the price he paid and made

Answer: 8

Step-by-step explanation:

64 ÷[4* 27 (-5^2

-5 times -5 =25

27-25=2

4*2=8

64 divided by 8= 8

8 is your answer

Answer:

It is not factorable

Step-by-step explanation:

The expression is not factorable with rational numbers.

Answer:

This is not factorable.

Step-by-step explanation:

This is not factorable.