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3 years ago

I need to somebody show work for number 13 please

Mathematics

1 answer:

Damm [24]3 years ago

6 0

<h3>Given</h3>

PQ = QR . . . . . . Q is the midpoint of PR
PQ = 3x+14
QR = 7x-10

<h3>Solution</h3>

Put the given expressions in the given relation.

... 3x +14 = 7x -10

Subtract the left side.

... 0 = (7x -10) -(3x +14)

... 0 = (7-3)x +(-10-14) . . . . . simplify

... 0 = 4x -24

Divide by the coefficient of x.

... 0 = x - 6

Add the opposite of the constant.

... 6 = x

The value of x is 6.

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A community center is serving a free meal to senior citizens. The center plans to feed 700 people in 4 hours. Write and solve an

Arada [10]

A. If the center plans to feed 700 people in 4 hours, then in order to find the average number of people the center is planning to feed each hour is 700/4 which is equal 175. So, the center is planning to feed 175 people per hour.

B. If during the first hour and a half, the center fed 270 people, find the remaining people to feed left: 700 - 270 = 430 people left to feed.

Please mark as brainliest

8 0

3 years ago

A TV and Washing machine were purchased for Rs 15000 each. If together, they are sold at Rs 35000, What is Profit and Profit%

Citrus2011 [14]

Given:

A TV and Washing machine were purchased for Rs 15000 each.

Together, they are sold at Rs 35000.

To find:

The profit and profit%.

Solution:

A TV and Washing machine were purchased for Rs 15000 each. So, the total cost price is

$C.P.=15000+15000$

$C.P.=30000$

Together, they are sold at Rs 35000. So, the selling price S.P. is 35000.

Now,

$Profit=S.P.-C.P.$

$Profit=35000-30000$

$Profit=5000$

And,

$Profit\%=\dfrac{Profit}{C.P.}\times 100$

$Profit\%=\dfrac{5000}{30000}\times 100$

$Profit\%=\dfrac{50}{3}$

$Profit\%\approx 16.67$

Therefore, the profit is Rs. 5000 and the profit percent is 16.67%.

5 0

3 years ago

A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a le

CaHeK987 [17]

Answer:

Step-by-step explanation:

Hello!

To compete in the touch screen phone market a manufacturer aims to release a new touch screen with a battery life said to last more than two hours longer than the leading product which is the desired feature in phones.

To test this claim two samples were taken:

Sample 1

X: battery lifespan of a unit of the new product (min)

n= 93 units of the new product

mean battery life X[bar]= 8:53hs= 533min

S= 84 min

Sample 2

X: battery lifespan of a unit of the leading product (min)

n= 102 units of the leading product

mean battery life X[bar]= 5:40 hs = 340min

S= 93 min

The population variances of both variances are unknown and distinct.

To test if the average battery life of the new product is greater than the average battery life of the leading product by 2 hs (or 120 min) the parameters of interest will be the two population means and we will test their difference, the hypotheses are:

H₀: μ₁ - μ₂ ≤ 120

H₁: μ₁ - μ₂ > 120

Considering that there is not enough information about the distribution of both variables, but both samples are big enough, we can apply the central limit theorem and approximate the distribution of both sample means to normal, this way we can use the standard normal:

$Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} } }$