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3 years ago

Help me in this question

Mathematics

1 answer:

Nat2105 [25]3 years ago

8 0

Answer:

Step-by-step explanation:

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A gym charges membership dues of $25 per month. Which equation represents the total cost, C, of belonging to the gym for m month

Anestetic [448]

Answer:

The answer is B.

Step-by-step explanation:

Given that the fixes amount is $25 per month. In order to find the total cost, C, you have to multiply $25 with the number of months.

C = 25m

6 0

3 years ago

Read 2 more answers

What is the exact value of sin -840

dsp73

First subtract 2 revolutions from the - 840 :- (- (840-720) = -120

This gives sin -120 which is in the 3rd quadrant of the unit circle

sin -120 = - sin 60 = - sqrt3/2

7 0

4 years ago

The difference of 28 and a number m is 3

jonny [76]

Answer:

28 - m = 3

28 - 3 = m

25 = m

Hope this helps :) mark me brainliest if you want to

8 0

3 years ago

Hi so could anyone please tell me how to solve this I’ve tried looking through my notes but I can’t find a question similar to t

neonofarm [45]

Answer:

no solution

Step-by-step explanation:

There are no values of x that make the equation true.

6 0

3 years ago

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

3 years ago