Write the inequality that represents the solution set for −6z17<−23

Which is the correct answer

Arturiano [62]

Answer:

I think it is -34 but i might not be right

Step-by-step explanation:

4 0

3 years ago

How do you do number 1?<br>Whenever I tried to answer it, I always get fraction. help me.

Flura [38]

Answer:

The pairs are (13,15) and (-15,-13).

Step-by-step explanation:

If n is an odd integer, the very next odd integer will be n+2.

n+1 is even (so we aren't using this number)

The sum of the squares of (n) and (n+2) is 394.

This means

(n)^2+(n+2)^2=394

n^2+(n+2)(n+2)=394

n^2+n^2+4n+4=394 since (a+b)(a+b)=a^2+2ab+b^2

Combine like terms:

2n^2+4n+4=394

Subtract 394 on both sides:

2n^2+4n-390=0

Divide both sides by 2:

n^2+2n-195=0

Now we need to find two numbers that multiply to be -195 and add up to be 2.

15 and -13 since 15(-13)=-195 and 15+(-13)=2

So the factored form is

(n+15)(n-13)=0

This means we have n+15=0 and n-13=0 to solve.

n+15=0

Subtract 15 on both sides:

n=-15

n-13=0

Add 13 on both sides:

n=13

So if n=13 , then n+2=15.

If n=-15, then n+2=-13.

Let's check both results

(n,n+2)=(13,15)

13^2+15^2=169+225=394. So (13,15) looks good!

(n,n+2)=(-15,-13)

(-15)^2+(-13)^2=225+169=394. So (-15,-13) looks good!

6 0

3 years ago

What is 8+7times15 so what is it tell me now

Inessa05 [86]

The answer is 113. As said by the calculator

5 0

3 years ago

Find the area A of triangle JKL with the vertices J(5,8), K(0,7), and L(5,4)

eimsori [14]

Answer:

Area of the triangle is 10 square units

Step-by-step explanation:

Recall that the formula for the area of a triangle is: "base x height / 2", and in this case, we can consider the triangle's base as the segment that joins the vertices (5, 4) and (5, 8) which gives a segment length of 4 units. The height of the triangle is then the distance between the third vertex (0,7) and the base, which is exactly 5 units. Then the area becomes:

Area = 4 x 5 / 2 = 10 square units.

5 0

3 years ago

A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 174 yellow peas. Use a 0.01

slavikrds [6]

Answer:

a) $z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

b) For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

Step-by-step explanation:

Data given and notation

n=420+174=594 represent the random sample taken

X=174 represent the number of yellow peas

$\hat p=\frac{174}{594}=0.293$ estimated proportion of yellow peas

$p_o=0.23$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of yellow peas is 0.23:

Null hypothesis: $p=0.23$

Alternative hypothesis: $p \neq 0.23$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z>3.649)=0.00026$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) Critical value

For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

5 0

3 years ago