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3 years ago

For the best set summarized in the boxplot, identify the interquartile range (IQR)

Mathematics

1 answer:

djverab [1.8K]3 years ago

6 0

<h3>Answer: 22 (choice C)</h3>

Work Shown:

IQR = Q3 - Q1

IQR = 70 - 48

IQR = 22

Note: Q1 and Q3 are the left and right edges of the box. The IQR is the width of the box.

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Divide £320 in the ratio 3:5

melamori03 [73]

Answer:

120:200

Step-by-step explanation:

3/8*320= 120

320-120=200

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3 years ago

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What is the area of ABC?

notsponge [240]

Let D be the Intersection of Height of the Triangle and Base of the Triangle BC

From the Figure, We can notice that Triangle ADB is a Right angled Triangle.

We know that, In a Right angled Triangle :

$\bigstar$ (Hypotenuse)² = (First Leg)² + (Second leg)²

In Triangle, ADB : AB is the Hypotenuse, AD is the First leg and BD is the Second leg

Given : AB = 15 and BD = 9

Substituting the values, We get :

$:\implies$ (15)² = (AD)² + (9)²

$:\implies$ 225 = (AD)² + 81

$:\implies$ (AD)² = 225 - 81

$:\implies$ (AD)² = 144

$:\implies$ (AD)² = (12)²

$:\implies$ AD = 12

We know that, In a Right angled Triangle :

$\bigstar\;\;\boxed{\mathsf{Tan\theta = \dfrac{Opposite\;Side}{Adjacent\;Side}}}$

Now, Consider Triangle ADC : With respect to 45° Angle, AD is the Opposite Side and DC is the Adjacent Side

$:\implies \mathsf{In\;Triangle\;ADC,\;Tan45^{\circ} = \dfrac{AD}{DC}}$

$\mathsf{:\implies 1 = \dfrac{12}{DC}}$

$:\implies$ DC = 12

$:\implies$ Total Length of the Base (BC) = BD + DC

$:\implies$ Total Length of the Base (BC) = 9 + 12

$:\implies$ Total Length of the Base (BC) = 21

We know that, Area of the Triangle is given by :

$\bigstar\;\;\boxed{\mathsf{Area = \dfrac{1}{2} \times Base \times Height}}$

In Triangle, ABC : AD is the Height and BC is the Base

$:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = \dfrac{1}{2} \times BC \times AD}$

$:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = \dfrac{1}{2} \times 21 \times 12}$

$:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = (21 \times 6)}$

$:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = 126}$

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4 years ago

Suppose that y varies directly with x, and y=4 when x=20. find y when x=9

Llana [10]

Answer:

y = 1.8

Step-by-step explanation:

y=kx

4=k(20)

k= $\frac{1}{5}$

When x = 9:

$y=\frac{1}{5} x=\frac{1}{5} (9)\\y=(\frac{9}{5}) = 1.8$

6 0

3 years ago

A magazine conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale,

Lyrx [107]

Answer:

a) The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

b) $SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

c) The 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

Step-by-step explanation:

Notation and previous concepts

$n_1 =35$ represent the sample of ships that carry fewer than 500 passengers

$n_2 =44$ represent the sample of ships that carry 500 or more passengers

$\bar x_1 =85.82$ represent the mean sample of of ships that carry fewer than 500 passengers

$\bar x_2 =81.90$ represent the mean sample of of ships that carry 500 or more passengers

$\sigma_1 =4.55$ represent the population deviation of ships that carry fewer than 500 passengers

$\sigma_2 =3.97$ represent the sample deviation of ships that carry 500 or more passengers

$\alpha=0.05$ represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$ (1)

Part a

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

Part b: At 95% confidence, what is the margin of error?

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

The standard error is given by the following formula:

$SE=\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$

And replacing we have:

$SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

Part c: What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships?

Confidence interval

Now we have everything in order to replace into formula (1):

$3.92-1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=2.009$

$3.92+1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=5.830$

So on this case the 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

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3 years ago

There are gallon of water in a 3-gallon container. What fraction of the<br> container is filled?

Darina [25.2K]

Answer:

1/3

Step-by-step explanation:

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