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aleksley [76]
3 years ago
5

) Kira drew a circle with a radius of 20 inches and then another circle with a radius of 10 inches. What is the approximate diff

erence between the areas of these two circles?
Mathematics
2 answers:
olga2289 [7]3 years ago
8 0

Answer:942.48 aprox

Step-by-step explanation:el área del primero es 1256.64 del segundo es 314.16 y si lo restamos da 942.48

La formula del área es rxrxπ

spin [16.1K]3 years ago
5 0

Answer:

10 inches

Step-by-step explanation:

Because if you draw a circle and the inside you draw another circle the spaces between then is 10 inches cause 20 - 10= 10

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Write an algebraic expression for two fifths the cube of a number
pashok25 [27]
N - the number

\dfrac{2}{5}n^3
7 0
3 years ago
Look at pseudocode segment below. y ← x + x + x + x + x // line a z ← y + y + y // line b If x = 6 before this segment is execut
r-ruslan [8.4K]

Answer:

90.

Step-by-step explanation:

We have two lines of code:

y ← x + x + x + x + x

z ← y + y + y

and, x=6 before the execution. Then, when the program starts to run we obtain:

y ← 6 + 6 + 6 + 6 + 6

y ← 30

z ← 30+30+30

z ← 90.

Then, the value of z after excecution is 90.

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3 years ago
Assume, for the sake of this question, that the data were collected through a well-designed, well-implemented random sampling me
amid [387]

Answer:

Since the pvalue of the test is 0.2743 > 0.1, the threshold probably was met.

Step-by-step explanation:

The widget manufacturing company had established a threshold of 60% preferring the proposed new widget to move forward with producing the new widgets.

This means that at the null hypothesis we test if the proportion is at least 60%, that is:

H_{0}: p \geq 0.6

And the alternate hypothesis is:

H_{a}: p < 0.6

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.6 is tested at the null hypothesis:

This means that:

\mu = 0.6

\sigma = \sqrt{0.6*0.4}

Three hundred thirty-eight of 575 respondents reported preferring the proposed new widget.

This means that n = 575, X = \frac{338}{575} = 0.5878

Value of the test-statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.5878 - 0.6}{\frac{\sqrt{0.6*0.4}}{\sqrt{575}}}

z = -0.6

Pvalue of the test and decision:

We want to find the probability of a proportion of 0.5878 or lower, which is the pvalue of z = -0.6.

Looking at the z-table, z = -0.6 has a pvalue of 0.2743.

Since 0.2743 > 0.1, the threshold probably was met.

3 0
3 years ago
Kristie has taken five tests in science class. The average of all five of Kristie's test scores is 94. The average of her last t
Bingel [31]

Answer:  The average of the first two test scores is 97.

Step-by-step explanation:  Given that Kristie has taken five tests in science class. The average of all five of Kristie's test scores is 94 and the average of her last three test scores is 92.

We are to find the average score of her first two tests.

Let a1, a2, a3, a4 and a5 be teh scores of Kristle in first , second, third, fourth and fifth tests respectively.

Then, according to the given information, we have

\dfrac{a1+a2+a3+a4+a5}{5}=94\\\\\Rightarrow a1+a2+a3+a4+a5=94\times5\\\\\Rightarrow a1+a2+a3+a4+a5=470~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

\dfrac{a3+a4+a5}{3}=92\\\\\Rightarrow a3+a4+a5=92\times3\\\\\Rightarrow a3+a4+a5=276~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Subtracting equation (ii) from equation (i), we get

(a1+a2+a3+a4+a5)-(a3+a4+a5)=470-276\\\\\Rightarrow a1+a2=194\\\\\Rightarrow \dfrac{a1+a2}{2}=\dfrac{194}{2}\\\\\Rightarrow \dfrac{a1+a2}{2}=97.

Thus, the average of the first two test scores is 97.

6 0
3 years ago
Kiran’s backpack weighs 3 pounds less than Clare’s backpack. Clare’s backpack weighs 14 pounds. How much does Kiran’s backpack w
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14-3=11, The answer is 11
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2 years ago
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