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3 years ago

-2s(5st+3t) need help with problem

1 answer:

7 0

The answer to this problem is -10x^2t-6st to get this answer or this problem you must use the product rule formula x^a x^b=x^a=b

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Solve for a a^2+17=42?

grandymaker [24]

We are given the following equation:

$a^2+17=42$

Subtract both sides by 17

$a^2=25$

Take the positive/negative square root of both sides

$a=\pm5$

This should be your answer. Let me know if you have any questions, thanks!

6 0

3 years ago

How do you find the mean of a stem and leaf plot

bonufazy [111]

Well, I know to find the mean you add all the numbers and divide it by how many numbers there were, so I'm guessing you just do that with the stem and leaf plot. I hoped this helped a bit

6 0

3 years ago

What is the value of the function at x=−2?

Nezavi [6.7K]

Answer:

I would think y equals 2

Step-by-step explanation:

If I'm wrong someone tell me

6 0

3 years ago

Read 2 more answers

Can someone help with this problem

WARRIOR [948]

35+25=60
y=60

If you add the opposite side values of the triangle (not the one touching the y), you get the value of y). 35+25=y
This works because y+x= 180 and the three angles of a triangle add to 180.

4 0

3 years ago

Read 2 more answers

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is:

$74.2-(1.8331)\frac{5.3083}{\sqrt{10} }$ ≤ μ ≤ $74.2+(1.8331)\frac{5.3083}{\sqrt{10} }$

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0

3 years ago