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3 years ago

Two groups of students were asked how many pets they had. The table below shows the numbers for each group:

Mathematics

2 answers:

Nana76 [90]3 years ago

5 0

where are the answers?

VLD [36.1K]3 years ago

4 0

Answer:

The Answer choices are this below

<em>The interquartile range for Group A students is 0.5 less than the interquartile range for Group B students. </em>

<em> The interquartile range for Group A students is 1.5 more than the interquartile range for Group B students. </em>

<em> The interquartile range for Group A students is equal to the interquartile range for Group B students. </em>

<em> The interquartile range for Group A students is 1 more than the interquartile range for Group B students.</em>

Step-by-step explanation:

The answer is d because it has a wider range than company b but i could be wrong this kind of math is not my specialty.

jorelis

3 years ago

thank you so much

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The failure of a circuit board interrupts work that utilizes a computing system until a new board is delivered. The delivery tim

Hatshy [7]

Answer:

The expected cost is 152

Step-by-step explanation:

Recall that since Y is uniformly distributed over the interval [1,5] we have the following probability density function for Y

$f_Y(y ) = \frac{1}{5-1} = \frac{1}{4}$ if $1\leq y \leq 5$ and 0 othewise. (To check this is the pdf, check the definition of an uniform random variable)

Recall that, by definition

$E(Y^k) = \int_{-\infty}^{\infty} y^kf_Y(y)dy$

Also, we are given that $C = 50+3Y+9Y^2$ . Recall the following properties of the expected value. If X,Y are random variables, then

$E(a+bX+cY) = a+bE(X)+cE(Y)$

Then, using this property we have that $E[C] = 50+3E[Y]+ 9E[Y^2]$ .

Thus, we must calculate E[Y] and E[Y^2].

Using the definition, we get that

$E[Y] = \int_{1}^{5}\frac{y}{4} dy =\frac{1}{4}\left\frac{y^2}{2}\right|_{1}^{5} = \frac{25}{8}-\frac{1}{8} = 3$

$E[Y^2] = \int_{1}^{5}\frac{y^2}{4} dy =\frac{1}{4}\left\frac{y^3}{3}\right|_{1}^{5} = \frac{125}{12}-\frac{1}{12} = \frac{31}{3}$

Then

$E(C) = 50 + 3\cdot 3 + 9 \cdot \frac{31}{3} = 152$

5 0

3 years ago

A random sample of 10161016 adults in a certain large country was asked "Do you pretty much think televisions are a necessity o

Korolek [52]

Answer:

0.509

Step-by-step explanation:

Given that a random sample of 1016 adults in a certain large country was asked "Do you pretty much think televisions are a necessity or a luxury you could do without?"

Of the 1016 adults surveyed, 517 indicated that televisions are a luxury they could do without.

To find out the point estimate of population proportion of adults in the country who believe that televisions are a luxury they could do without.

We find that point estimate is nothing but sample proportion

Sample proportion = favourably voters/total people surveyed

= $\frac{517}{1016} \\=0.5088\\=0.509$

6 0

3 years ago

Can someone please help me with these questions? I’m desperate! As always, I will give brainliest . I promise.

professor190 [17]

Since the centre of dilatation is the origin, we only require to multiply each of the coordinate points by the scale factor 3

(- 2, 1 ) → ( - 6, 3 )

(- 2, 3 ) → (- 6, 9 )

(1, 2 ) → (3, 6 )

(1, 3 ) → (3, 9 )

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3 years ago

The problem you chose:<br> 2x2+2x-12 = 0

dangina [55]

x = 4

Simplifying the equation...

2(2) + 2x - 12 = 0, 2x - 8 = 0, x = 4

7 0

3 years ago

A house that is built for $200,000 and then marked up 175%<br> cost+(cost x % of increase)

Gekata [30.6K]

Answer:

ФΔHello wonderful personΔФ

Your answer should be:

$235,000

$200,000×0.175 = 35,000+ 200,000

Step-by-step explanation:

<h2>First answer!!!</h2><h3>Please mark brainliest≅</h3><h3>Have a great day!</h3>

3 0

3 years ago

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