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2 years ago

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Sara threw her math book off a building . The height of the book, seconds after being thrownis modeled by the equation below. Th

e height measured in feet. h(x)=-5(x+2)(x-7) what is the height of the book when it is thrown?

1 answer:

Vikentia [17]2 years ago

4 0

Answer: 70 ft

Step-by-step explanation:

Given

height of book is measured by the function is

$h(x)=-5(x+2)(x-7)$

when the book is thrown $x=0$

$\therefore h(x=0)=-5(0+2)(0-7)\\\Rightarrow h(x=0)=70\ ft$

The height of the book is $70\ ft$

You might be interested in

I need help asap!! before 2:25 please and thank youu!!

Lena [83]

Answer:

B. (2, -1)

Step-by-step explanation:

Take tracing paper and mark the origin and point m. Rotate 90 degrees clockwise around the origin. Point m will be at (2, -1).

7 0

3 years ago

A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed i

Hunter-Best [27]

Answer:

a) σ/√n= 1.43 min

c) Margin of error 2.8028min

d) [30.1972; 35.8028]min

e) n=62 customers

Step-by-step explanation:

Hello!

The variable of interest is

X: Time a customer stays at a restaurant. (min)

A sample of 49 lunch customers was taken at a restaurant obtaining

X[bar]= 33 mi

The population standard deviation is known to be δ= 10min

a) and b)

There is no information about the distribution of the population, but we know that if the sample is large enough, n≥30, we can apply the central limit theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;σ²/n)

Where μ is the population mean and σ²/n is the population variance of the sampling distribution.

The standard deviation of the mean is the square root of its variance:

√(σ²/n)= σ/√n= 10/√49= 10/7= 1.428≅ 1.43min

c)

The CI for the population mean has the general structure "Point estimator" ± "Margin of error"

Considering that we approximated the sampling distribution to normal and the standard deviation is known, the statistic to use to estimate the population mean is Z= (X[bar]-μ)/(σ/√n)≈N(0;1)

The formula for the interval is:

[X[bar]± $Z_{1-\alpha /2}$ *(σ/√n)]

The margin of error of the 95% interval is:

$Z_{1-\alpha /2}= Z_{1-0.025}= Z_{0.975}= 1.96$

d= $Z_{1-\alpha /2}$ *(σ/√n)= 1.96* 1.43= 2.8028

d)

[X[bar]± $Z_{1-\alpha /2}$ *(σ/√n)]

[33±2.8028]

[30.1972; 35.8028]min

Using a confidence level of 95% you'd expect that the interval [30.1972; 35.8028]min contains the true average of time the customers spend at the restaurant.

e)

Considering the margin of error d=2.5min and the confidence level 95% you have to calculate the corresponding sample size to estimate the population mean. To do so you have to clear the value of n from the expression:

d= $Z_{1-\alpha /2}$ *(σ/√n)

$\frac{d}{Z_{1-\alpha /2}}$ = σ/√n

√n*( $\frac{d}{Z_{1-\alpha /2}}$ )= σ

√n= σ* ( $\frac{Z_{1-\alpha /2}}{d}$ )

n=( σ* ( $\frac{Z_{1-\alpha /2}}{d}$ ))²

n= (10* $\frac{1.96}{2.5}$ )²= 61.47≅ 62 customers

I hope this helps!

3 0

3 years ago

Add the fractions and simplify 5 2/5 + 2 10/13=

HACTEHA [7]

Simple....

you have:

$5 \frac{2}{5} +2 \frac{10}{13}$

to convert them into improper fractions:

1.) Take whole number*denominator

(5*5=25)

2.) Take what you got...and add the numerator

(25+2=27)

3.) Use the original denominator

$\frac{27}{5}$

Do the same to the other one...

2*13=26

26+10=36

$\frac{36}{13}$

Now..add...

$\frac{27}{5} + \frac{36}{13}$

But..you need a common denominator!

$\frac{27}{5} * \frac{13}{13}= \frac{351}{65}$

$\frac{36}{13} * \frac{5}{5} = \frac{180}{65}$

Now you can add!

$\frac{351}{65} + \frac{180}{65} = \frac{531}{65}$

When simplified....

$8 \frac{11}{65}$

Thus, your answer.

6 0

3 years ago

Duane And his family are playing a game Dwayne scores 11 points each tile matches 1/3 point model the solution with an equation

Luba_88 [7]

Answer:

33 tiles

Step-by-step explanation:

Given

$1\ tile = \frac{1}{3}\ points$

Required

Determine the number of tiles in 11 points

Represent this with x. So, we have:

$1\ tile = \frac{1}{3}\ points$

$x = 11\ points$

Cross Multiply

$x * \frac{1}{3} = 11 * 1$

Multiply through by 3

$3*x * \frac{1}{3} = 11 * 1*3$

$x = 33$

6 0

3 years ago

A music school has budgeted to purchase 3 musical instruments. They plan to purchase a piano costing $4,000, a guitar costing $6

kondaur [170]

Answer:

1) $z=\frac{4000-4500}{2500}=-0.2$

That means 0.2 deviations below the mean

2) $z=\frac{600-500}{200}=0.5$

That means 0.5 deviations above the mean

3) $z=\frac{750-850}{100}=-1$

That means 1 deviations below the mean

4) For this case the lowest instrument compared to the other instruments of the same type is the drum set since have the lowest z score z =-1

5) For this case the highest instrument compared to the other instruments of the same type is guitar since have the highest z score z =0.5

Step-by-step explanation:

They plan to purchase a piano costing $4,000, a guitar costing $600, and a drum set costing $750.

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part 1

Let X the random variable that represent the cost of for a piano, and for this case we know this

Where $\mu=4500$ and $\sigma=2500$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we replace the formula for X= 4000 we got:

$z=\frac{4000-4500}{2500}=-0.2$

That means 0.2 deviations below the mean

Part 2

Let X the random variable that represent the cost of for a guitar, and for this case we know this

Where $\mu=500$ and $\sigma=200$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we replace the formula for X= 600 we got:

$z=\frac{600-500}{200}=0.5$

That means 0.5 deviations above the mean

Part 3

Let X the random variable that represent the cost of for drum set, and for this case we know this

Where $\mu=850$ and $\sigma=100$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we replace the formula for X= 750 we got:

$z=\frac{750-850}{100}=-1$

That means 1 deviations below the mean

Part 4

For this case the lowest instrument compared to the other instruments of the same type is the drum set since have the lowest z score z =-1

Part 5

For this case the highest instrument compared to the other instruments of the same type is guitar since have the highest z score z =0.5

4 0

3 years ago