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3 years ago

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The heights of the children at a summer camp are normally distributed with a mean of 54 in. and a standard deviation of 2.4 in.

Enter the height of a child, whose z-score is 1.5, in the box. Round your answer to the nearest tenth.

1 answer:

Andreyy893 years ago

6 0

The z-score is a measure of how many standard deviations the value is from the mean. A z-score of 1.5 means that the value is 1.5 standard deviations above the mean.

<span>1.5 x 2.4 = 3.6 </span>
<span>54 + 3.6 = 57.6 </span>

<span>Answer: </span>
<span>57.6 in </span>

<span>The formula is: </span>
<span>x = zσ + μ </span>

<span>z : z-score (1.5) </span>
<span>σ : standard deviation (2.4) </span>
<span>μ : mean (54) </span>

<span>Answer: </span>
<span>x = 57.6</span>

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Sam is planting a garden that measures

nikitadnepr [17]

Answer:

63.7 feet

Step-by-step explanation:

We can re-write the width and the length both as rational numbers:

- Width: $w=6.5=\frac{65}{10}$

- Length: $L=9.8=\frac{98}{10}$

So now we can find the area of the garden by multiplying width and length:

$A=wL=\frac{65}{10}\cdot \frac{98}{10}=\frac{65\cdot 98}{10\cdot 10}=\frac{6370}{100}=63.7$

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3 years ago

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

or

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

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3 years ago

Cybil flips a coin and rolls a fair number cube at the same time. What is the probability that she will toss tails and roll a nu

Firdavs [7]

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3 years ago

How do you solve his with working

AlexFokin [52]

Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

$\bf \begin{array}{cllll}
\textit{circumference of a circle}\\\\ 
2\pi r
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{arc's length}\\\\
s=\cfrac{\theta r\pi }{180}
\end{array}\\\\
-------------------------------$

$\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+
\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}
\\\\\\
15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888$

b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.

$\bf \begin{array}{cllll}
\textit{area of a circle}\\\\ 
\pi r^2
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{area of a sector of a circle}\\\\
s=\cfrac{\theta r^2\pi }{360}
\end{array}\\\\
-------------------------------$

$\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}
\\\\\\
90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884$

7 0

3 years ago

A projector was sold after allowing 10% discount on the marked price and levying 13% VAT .If the selling price of the projector

Tom [10]

The marked price of the projector is $50,000.

<h3>What is the marked price?</h3>

The price of the projector after the discount can be represented with:

(100 - 10%)x - 90%x = 0.90x

Where x is the marked price

The price of the projector after the VAT is : (1.13) x 0.90x = 1.02x

Difference in price = 1.02x - 0.90x = 5850

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x = 5850 / 0.12

x = $50,000

To learn more about how to calculate discounts, please check: brainly.com/question/26061308

#SPJ1

8 0

2 years ago