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IgorLugansk [536]

2 years ago

7

5 friends go out to lunch and order 4 pizzas . The friends divided the pizzas evenly. find how much pizza each friend got as a d

ecimal

1 answer:

Ainat [17]2 years ago

8 0

0.8 I think. I'm not 100% sure

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Determine the intercepts

SpyIntel [72]

is there more ???????

6 0

2 years ago

a store marks up the merchandise it sells by 40%. If the regular price of the item WAS $25 how much is the store selling it for.

devlian [24]

So if the regular price of an item is 25 and the store is marking it up by 40%, then we have to multiply 25 by .4 to find out what 40% of 25 dollars is. 40% of 25 is 10 dollars. So now you just do 25 + 10 and you get 35. Answer: $35

7 0

2 years ago

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

2 years ago

Divide 2 1/3 divided by 2/5

anastassius [24]

Answer:5.8 or just 5

Step-by-step explanation: 35/6 = 5 5/6 = 5.8333333333

3 0

2 years ago

In a jar, there is a green, a blue, and a red marble. You draw (with replacement) a marble from the jar until you get the green

denpristay [2]

Answer:

The value is $P( X = m ) = (\frac{2}{3} )^{m-1 } * \frac{1}{3}$

Step-by-step explanation:

From question we are told that

The number of types of marbles present jar is n = 3

Generally the probability of drawing a green marble is

$P(g) = \frac{1}{3}$

Generally the probability of drawing a marble that is not green is

$P(g') = 1 - \frac{1}{3} = \frac{2}{3}$

From the question we are told that there will be continuous drawing of marbles from the jar (in such a way that after each marble is drawn it is being replaced) until a green marble drawn

Let m be the number of times marbles has been drawn when a green marble was gotten

it then means that for m - 1 times the marbles where drawn a green marble was not obtain.

Generally the probability drawing m times is mathematically is mathematically represented as

$P( X = m ) = (P(g'))^{m-1 } * P(g)$

=> $P( X = m ) = (\frac{2}{3} )^{m-1 } * \frac{1}{3}$

7 0

2 years ago