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1 year ago

The mean of four numbers is 59, The numbers are 52, 58, 57, and x. Find the value of x.

Mathematics

1 answer:

myrzilka [38]1 year ago

4 0

Answer:

X=69

Step-by-step explanation:

52, 58, 57, X =59

52+58+57+X/4=59

167+X/4=59

Multiply both sides by 4

4(167+X) /4 = 59x4

Simplify

167 + X= 236

Subtract 167 from both sides

167 + X- 167 = 236 – 167

Simplify

X= 69

to check

52+ 58+57+69/4 =59

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A student has 10 coins in her pocket that have a value of $0.85. The coins are either

il63 [147K]

Answer:

$\left\{\begin{matrix}x+y=10\\ 0.05x+0.10y=0.85\end{matrix}\right.$

Step-by-step explanation:

System of Equations

Let's call:

x = number of nickels in the student's pocket

y = number of dimes in the student's pocket

Each nickel has a value of $0.05, so x nickels have a value of 0.05x

Each dime has a value of $0.10, so y dimes have a value of 0.10y

The student has a total of 10 coins, thus:

$x+y=10$

The total value of the coins is $0.85, thus

$0.05x+0.10y=0.85$

The system of linear equations that represents this scenario is:

$\left\{\begin{matrix}x+y=10\\ 0.05x+0.10y=0.85\end{matrix}\right.$

5 0

2 years ago

Someone help me with this ASAP!!!

hjlf

Answer:

95 over 2

because if it is 47 you double that since the denominator is 2 then you add the numerator and now you have your answer.

3 0

3 years ago

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I Need Help!!

andre [41]

There is no local extremas but here is the objective function graphed

7 0

2 years ago

A simple random sample of items resulted in a sample mean of . The population standard deviation is . a. Compute the confidence

Varvara68 [4.7K]

Answer:

(a): The 95% confidence interval is (46.4, 53.6)

(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

$n = 30$ -- sample size

$\bar x = 50$ -- sample mean

$\sigma = 10$ --- sample standard deviation

Solving (a): The confidence interval of the population mean

Calculate the standard error

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {30}}$

$\sigma_x = \frac{10}{5.478}$

$\sigma_x = 1.825$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.825$

$E = 3.577$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 3.577$

$Lower = 46.423$

$Lower = 46.4$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 3.577$

$Upper = 53.577$

$Upper = 53.6$ --- approximated

So, the 95% confidence interval is (46.4, 53.6)

Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {90}}$

$\sigma_x = \frac{10}{9.49}$

$\sigma_x = 1.054$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.054$

$E = 2.06584$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 2.06584$

$Lower = 47.93416$

$Lower = 47.9$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 2.06584$

$Upper = 52.06584$

$Upper = 52.1$ --- approximated

So, the 95% confidence interval is (47.9, 52.1)

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

$n = 30$ $E = 3.577$

In (b), we have:

$n = 90$ $E = 2.06584$

Notice that the margin of error decreases when the sample size increases.

4 0

3 years ago

The ratio of the area to the circumference is A. 2r B.r/2 C. 2/r

weqwewe [10]

It's nothing because e=mc^2

4 0

3 years ago

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