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3 years ago

13

The amount of jen's monthly phone bill is normally distributed with a mean of $65 and a standard deviation of 11. what percentag

e of her phone bills are between 32 and 98

1 answer:

stepan [7]3 years ago

4 0

It would be 99%. If you draw the normal distribution curve and plot the mean and standard deviation, it might help you visually.

:)

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Pls help!!!! <br>Evaluate the permutations <br><br><br>1. 6P2<br><br>2. 7P4<br><br>3. 8P3<br><br>

Gnom [1K]

Answer:

5p9

Step-by-step explanation:

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3 years ago

Anyone want to talk?

Tamiku [17]

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2 years ago

Miyoko's goal is to earn more than \$950$950dollar sign, 950 this week. She earns \$250$250dollar sign, 250 for every day she wo

GrogVix [38]

$250 c+ $ 180 g > $ 950

<u>Step-by-step explanation:</u>

As a cryptographer (c), Miyoko earns per day = $ 250

As a geologist (g) , Miyoko earns per day = $ 180

So the equation comes to be $250 c+ $ 180 g = $ 950

The equation can be rewritten to find c as, (950-180 g) / 250

The equation can be rewritten to find g as, (950 - 250 c) / 180

Plugin different values of c and g in the above 2 equations, we can find that ,

To achieve the goal, Miyoko requires to be a geologist for 3 days and crpytographist for 2 days.

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3 years ago

Identify this conic section. 16y = x^2

kiruha [24]

ANSWER

A parabola.

EXPLANATION

The given conic is :

$16y = {x}^{2}$

This can be rewritten as:

${x}^{2} = 16y$

${x}^{2} = 4(4)y$

This is a parabola with the vertex at the origin.

The foci is (0,4)

Therefore the given conic section is a parabola that has an axis of symmetry parallel to the y-axis.

7 0

3 years ago

Use identities to find the values of the sine and cosine functions for the following angle measure.

puteri [66]

Using the cosine double angle formula,

$\cos 2\theta=2\cos^2 \theta-1=\frac{12}{13}\\\\2\cos^{2} \theta=\frac{25}{13}\\\\\cos^{2} \theta=\frac{25}{26}\\\\\boxed{\cos \theta=\frac{5}{\sqrt{26}}}$

(Note I took the positive case since $\theta$ terminates in the first quadrant)

Using the Pythagorean identity,

$\sin^2 \theta+\cos^2 \theta=1\\\\\sin^2 \theta+\frac{25}{26}=1\\\\sin^2 \theta=\frac{1}{26}\\\\\boxed{\sin \theta=\frac{1}{\sqrt{26}}}$

(Note I took the positive case since $\theta$ terminates in the first quadrant)

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1 year ago