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3 years ago

9

A certain standardized test measures students’ knowledge in English and math. The English and math scores for 10 randomly select

ed students were recorded and analyzed. The results are shown in the computer output.
Which of the following represents the standard deviation of the residuals?

1.223

34.55

78.712

124.13

I think it's (B), 34.55

1 answer:

creativ13 [48]3 years ago

7 0

Answer:

(B) 34.55

S is the standard deviance for residuals.

ED2021

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Find fractional notation 5.6%

Vesna [10]

$\begin{gathered} 5.6\text{\%=}\frac{5.6}{100} \\ \Rightarrow\frac{56}{10}\div100 \\ \Rightarrow\frac{56}{10}\times\frac{1}{100} \\ \Rightarrow\frac{56}{1000} \\ Divide\text{ both numerator and denominator by 8, we have:} \\ \frac{7}{125} \end{gathered}$

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11 months ago

Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = 3 (*sqaure root sign*) t (20 − t), [0,

mafiozo [28]

9514 1404 393

Answer:

maximum: 15∛5 ≈ 25.6496392002
minimum: 0

Step-by-step explanation:

The minimum will be found at the ends of the interval, where f(t) = 0.

The maximum is found in the middle of the interval, where f'(t) = 0.

$f(t)=\sqrt[3]{t}(20-t)\\\\f'(t)=\dfrac{20-t}{3\sqrt[3]{t^2}}-\sqrt[3]{t}=\sqrt[3]{t}\left(\dfrac{4(5-t)}{3t}\right)$

This derivative is zero when the numerator is zero, at t=5. The function is a maximum at that point. The value there is ...

f(5) = (∛5)(20-5) = 15∛5

The absolute maximum on the interval is 15∛5 at t=5.

5 0

2 years ago

Find two positive integers such that the sum of the first number and four times the second number is 1000, and the product of th

vladimir2022 [97]

Answer:

The two numbers are:

x = 500

y = 125

Step-by-step explanation:

We want to find two numbers x and y, such that:

x + 4*y = 1000

f(x, y) = x*y is maximum.

From the first equation, we can isolate one of the variables to get

x = 1000 - 4y

now we can replace it in f(x, y):

x*y = (1000 - 4*y)*y = 1000*y - 4*y^2

So now we want to maximize the function:

f(y) =- 4*y^2 + 1000*y

where y must be an integer.

Notice that this is a quadratic equation with a negative leading coefficient (so the arms of the graph will open downwards), thus, the maximum will be at the vertex.

Remember that for a general quadratic equation:

y = a*x^2 + bx + c

the x-value of the vertex is:

x = -b/(2*a)

so, in the case of:

f(y) =- 4*y^2 + 1000*y

the y-value of the vertex will be:

y = -1000/(2*-4) = 1000/8 = 125

So we found the value of y.

now we can use the equation:

x = 1000 - 4*y

x = 1000 - 4*125 = 1000 - 500 = 500

x = 500

Then the two numbers are:

x =500

y = 125

6 0

2 years ago

The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen

vfiekz [6]

Answer:

a) 0.2581

b) 0.4148

c) 17

Step-by-step explanation:

For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$p = 0.75$

a. If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X = 9)$ when $n = 12$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{12,9}.(0.75)^{9}.(0.25)^{3} = 0.2581$

b. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X \geq 16)$ when $n = 20$

So

$P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 16) = C_{20,16}.(0.75)^{16}.(0.25)^{4} = 0.1897$

$P(X = 17) = C_{20,17}.(0.75)^{17}.(0.25)^{3} = 0.1339$

$P(X = 18) = C_{20,18}.(0.75)^{18}.(0.25)^{2} = 0.0669$

$P(X = 19) = C_{20,19}.(0.75)^{19}.(0.25)^{1} = 0.0211$

$P(X = 20) = C_{20,20}.(0.75)^{20}.(0.25)^{0} = 0.0032$

So

$P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1897 + 0.1339 + 0.0669 + 0.0211 + 0.0032 = 0.4148$

c. If you call 22 times, what is the mean number of calls that are answered in less than 30 seconds? Round your answer to the nearest integer.

The expected value of the binomial distribution is:

$E(X) = np$

In this question, we have $n = 22$

So

$E(X) = 22*0.75 = 16.5$

The closest integer to 16.5 is 17.

7 0

3 years ago

If f(1)=10 and f(n)=-5f(n-1)-2 then find the value of f(4)

Flura [38]

Answer:

$\large\boxed{f(4)=-1292}$

Step-by-step explanation:

$f(n)=-5f(n-1)-2\\\\f(1)=10\\\\f(2)=-5f(2-1)-2=-5f(1)-2\\\text{put}\ f(1)=10:\\f(2)=-5(10)-2=-50-2=-52\\\\f(3)=-5f(3-1)-2=-5f(2)-2\\\text{put}\ f(2)=-52\\f(3)=-5(-52)-2=260-2=258\\\\f(4)=-5f(4-1)-2=-5f(3)-2\\\text{put}\ f(3)=258\\f(4)=-5(258)-2=-1290-2=-1292$

4 0

3 years ago