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3 years ago

5

Substitution y=1/2x+2, y=x+1

1 answer:

dem82 [27]3 years ago

6 0

Answer: point form (2,3) and equation form is x=2 y=3

Step-by-step explanation:

would you like for me to explain it for you

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Please answer as soon as possible!!! Thank you

True [87]

Answer:

5,0 is the answer of ur question

6 0

3 years ago

Which function has zeros of -8, 1, and 3?

rodikova [14]

Answer:

the answer should be A

Step-by-step explanation:

Hope this helps!

7 0

2 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

Korolek [52]

Answer:

A sample of 499 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

In this question, we have that:

$\pi = 0.21$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 90% confidence level with an error of at most 0.03

We need a sample of n, which is found when $M = 0.03$ . So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.645\sqrt{\frac{0.21*0.79}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.21*0.79}$

$\sqrt{n} = \frac{1.645\sqrt{0.21*0.79}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.645\sqrt{0.21*0.79}}{0.03})^2$

$n = 498.81$

Rounding up

A sample of 499 is needed.

8 0

3 years ago

Suppose you just received a shipment of ten televisions. two of the televisions are defective. if two televisions are randomly

amid [387]

Try this solution:

1. for probability of both televisions work:

$P_{2w}=\frac{C^2_8}{C^2_{10}}=\frac{7*8}{9*10}=0.6(2)$

2. for probability of at least on is defective (it means, the sum = one is defective and two are defective):

$P_{1d}=\frac{C^1_2C^1_8}{C^2_{10}}=\frac{4*8}{9*10}=0.3(5)$ - for P(1defective);

$P_{2d}=\frac{C^2_2}{C^2_{10}}=\frac{2}{9*10}=0.0(2)$ - for P(2defective);

totaly for at least one is defective:

$P_{at \ least \ 1 \ d}=P_{1d}+P_{2d}=\frac{16}{45} +\frac{1}{45} =0.3(7)$

5 0

3 years ago

Represent A Quadratic Function Using:

Pani-rosa [81]

Sub the Xs in the eqn

5 0

2 years ago