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3 years ago

Using the data in the table,what is the cost of 3 apples?

Mathematics

1 answer:

anastassius [24]3 years ago

4 0

Answer:you add anmd subtract

Step-by-step explanation:

the answer is 3 to 4

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What is 7x-6=9x+6? PLEASE HELP ME!!!!!!!!!!!!!

vovangra [49]

Answer:

x=-6

Step-by-step explanation:

7x-9x=6+6

-2x=12

x=-6

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3 years ago

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The difference between two numbers is 9, and the sum of the squares of each number is 153. What is the value of the product of t

Mandarinka [93]

Numbers are 3 and 12
The product is 36

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3 years ago

Enter the amplitude of the function f(x).<br><br> f(x)=−sin(3x)−1

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Enter the amplitude of the function f(x). f(x)=-2sin(3x)-1 . 2. See answers. Log in to add comment. Answer. Answer. 5.0/5. 5.0/5.

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2 years ago

WILL GIVE BRAINLIEST IMMEDIATELY!!!!!!! GIVE EXPLANATION + ANSWER FOR BRAINLIEST

Monica [59]

The correct answer is:

D. ∠FGE≅∠NMP

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3 years ago

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The USA Today reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amo

Aloiza [94]

Answer:

$a)\ \ \bar x_m-\bar x_f=67.03\\\\b)\ \ E=15.7416\\\\c)\ \ CI=[51.2884, \ 82.7716]$

Step-by-step explanation:

a. -Given that:

$n_m=41\ , \ \sigma_m=33, \bar x_m=135.67\\\\n_f=37. \ \ ,\sigma_f=20, \ \ \bar x_f=68.64$

#The point estimator of the difference between the population mean expenditure for males and the population mean expenditure for females is calculated as:

$\bar x_m-\bar x_f\\\\\therefore \bigtriangleup\bar x=135.67-68.64\\\\=67.03$

Hence, the pointer is estimator 67.03

b. The standard error of the point estimator, $\bar x_m-\bar x_f$ is calculated by the following following:

$\sigma_{\bar x_m-\bar x_f}=\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

-And the margin of error, E at a 99% confidence can be calculated as:

$E=z_{\alpha/2}\times \sigma_{\bar x_m-\bar x_f}\\\\\\=z_{0.005}\times\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\\\=2.575\times \sqrt{\frac{33^2}{41}+\frac{20^2}{37}}\\\\\\=15.7416$

Hence, the margin of error is 15.7416

c. The estimator confidence interval is calculated using the following formula:

$\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

#We substitute to solve for the confidence interval using the standard deviation and sample size values in a above:

$CI=\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\=(135.67-68.64)\pm 15.7416\\\\=67.03\pm 15.7416\\\\=[51.2884, \ 82.7716]$

Hence, the 99% confidence interval is [51.2884,82.7716]

7 0

3 years ago