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3 years ago

100 POINTS PRE-CALCULUS

Mathematics

2 answers:

kiruha [24]3 years ago

4 0

Answer:

The first one is the third option and the second one is B.

Step-by-step explanation: Did the assignment.

Sonja [21]3 years ago

3 0

Answer:

not sure for the 1st one but pretty sure 2nd one is B

Step-by-step explanation:

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Fill in the blank. _______ are sample values that lie very far away from the majority of the other sample values. Centers Distr

cestrela7 [59]

Answer:

Outliers are sample values that lie very far away from the majority of the other sample values. Centers Distributions Outliers Frequencies are sample values that lie very far away from the majority of the other sample values.

Step-by-step explanation:

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3 years ago

The area of the circle below is 28.26. What is the diameter?

svet-max [94.6K]

I think the answer would be 9 or 18 not really sure

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3 years ago

If a/b and c/d are rational expressions then a/b divided by c/d =a times d/b times c true or false

Tomtit [17]

Answer:

(a d)/(bc)

Step-by-step explanation:

a/b ÷ c/d

Copy dot flip

a/b * d/c

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3 years ago

Evaluate 2 ∙ (8 + 4) using the steps of order of operations.

Natalka [10]

Step-by-step explanation:

Follow PEMDAS

$2 ∙ (8 + 4)$

P: Parentheses

E : Exponents

M: Multiplication

D: Division

A: Addition

S: Subtraction

$2 \cdot \: (12) \\ = 2 \times 12 \\ 24$

Question 2

$15 + 6 \times 3 \\ 15 + (6 \times 3) \\ 15 + 18 \\ = 33$

7 0

3 years ago

Read 2 more answers

A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq

Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

$s=16.6$ represent the sample deviation

$\bar X=12.8$ represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

$\alpha=1-0.99=0.01$ represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

$\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$12.8 - 2.58 \frac{16.6}{\sqrt{1003}}=11.448$

$12.8 + 2.58 \frac{16.6}{\sqrt{1003}} =14.152$

And the 99% confidence interval would be given (11.448;14.152).

We are confident that about 11 to 14 are the number of books that the people had read on the last year on average, at 1% of significance.

3 0

3 years ago