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3 years ago

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The class scores of a history test have a normal distribution with a mean Mu = 79 and a standard deviation Sigma = 7. If Opal’s

test score was 72, which expression would she write to find the z-score of her test score?.

1 answer:

e-lub [12.9K]3 years ago

3 0

Using the normal distribution, we got that expression for the z-score of her test score is

$z= \frac{72-29}{7} \\\\z=-7/7\\\\z=-1$

<h3>What is Normal Probability Distribution?</h3>

it's a type of probability distribution which is symmetric about the mean, it gives data near the mean are more frequent than data far from the mean.

Given that mean = $\mu=79$

and standard deviation = $\sigma = 7$

X=73

we know that formula of z-score is

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

Hence Z-score can be calculated as

$z= \frac{72-29}{7} \\\\z=-7/7\\\\z=-1$

Using the normal distribution, we got that expression for the z-score of her test score is

$z= \frac{72-29}{7} \\\\z=-7/7\\\\z=-1$

To learn more about the normal distribution visit : brainly.com/question/24663213

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It is estimated that the annual cost of driving a certain new car is given by the formula C = 0.37m + 2600 where m represents th

arsen [322]

Answer:

12000-14000 (in miles)

Step-by-step explanation:

We are given that the annual cost of driving a certain car is given by the formula

$C=0.37m+2600$

Where m=Represents the number of miles driven per year

C=Cost in dollars

We have to find the range of miles that Jane can drive her new car.

Range of budgets=$7040-$7780

Substitute C=$7040 in the given formula

$7040=0.37m+2600$

$7040-2600=0.37m$

$4440=0.37m$

$m=\frac{4440}{0.37}$

$m=12000$

Substitute C=$7780 in the given formula

$7780=0.37m+2600$

$7780-2600=0.37m$

$5180=0.37m$

$m=\frac{5180}{0.37}=14000$

The range of miles that Jane can drive her new car=12000-14000

4 0

3 years ago

Can someone please explain how to solve this step by step with answer? Thank you.

lidiya [134]

a)

well, she put 4000, and she earned in interest 960, so her accumulated amount is just their sum, 4960.

b)

now, it doesn't say, so we're assuming is <u>simple interest</u>, as opposed to compound interest.

$\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill&960\\ P=\textit{original amount deposited}\dotfill & \$4000\\ r=rate\to r\%\to \frac{r}{100}\dotfill\\ t=years\dotfill &3 \end{cases} \\\\\\ 960=(4000)(r)(3)\implies \cfrac{960}{(4000)(3)}=r\implies 0.08=r \\\\\\ \stackrel{\textit{converting to percentage}}{r=0.08\cdot 100}\implies r=\stackrel{\%}{8}$

c)

let's make the rate 1% greater then, and check

$\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill\\ P=\textit{original amount deposited}\dotfill & \$4000\\ r=rate\to \stackrel{8+1}{9\%}\to \frac{9}{100}\dotfill&0.09\\ t=years\dotfill &3 \end{cases} \\\\\\ I=(4000)(0.09)(3)\implies I=1080 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{at 9\%}}{1080}-\stackrel{\textit{at 8\%}}{960}\implies \stackrel{\textit{that much more}}{120}$

4 0

3 years ago

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks

Fudgin [204]

Answer:

The teenagers spent 4.5 hours per week, on average, on the phone.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 4.5

Sample mean, $\bar{x}$ = 4.75

Sample size, n = 15

Alpha, α = 0.05

Sample standard deviation, s = 2

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 4.5\text{ hours per week}\\H_A: \mu > 4.5\text{ hours per week}$

We use one-tailed(right) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{4.75 - 4.5}{\frac{2}{\sqrt{15}} } = 0.4841$

Now,

$t_{critical} \text{ at 0.05 level of significance, 14 degree of freedom } = 1.76$

Since,

$t_{stat} < t_{critical}$

We accept the null hypothesis. Thus, the teenagers spent 4.5 hours per week, on average, on the phone. The sample contradicted the organization's claim.

7 0

4 years ago

Shrinivas deposited $1300 in an account which pays 8% annual interest, compounded continuously. How long will it take to reach $

sertanlavr [38]

Answer: a. 12.4 years

Step-by-step explanation:

Function to represent exponential growth continuously:

$A=A_0e^{rt}$ ...(i)

, where $A_0$ = initial amount, r= rate of interest ( in decimal) , A = Amount after t years.

As per given , we have

$A_0=\$1300\\\\r=8\%=0.08\\\\ A=\$3500$

To find : t

Put all values in (i) ,

$3500=(1300)e^{0.08t}\\\\\Rightarrow\ 2.6923077=e^{0.08t}$

Taking natural log on both sides

$\ln(2.6923076)= 0.08t\\\\\Rightarrow\ 0.99039867=0.08t\\\\\Rightarrow\ t=\dfrac{0.99039866}{0.08}=12.37998325\approx12.4$

Hence, it would take <u>12.4</u> years.

so, the correct option is a. 12.4 years .

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3 years ago

What’s the sum (3x – 5) + (-4x + 1)

MaRussiya [10]

Answer:

−x−4

Step-by-step explanation:

7 0

3 years ago

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