All categories

2 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Mathematics

1 answer:

blondinia [14]2 years ago

3 0

Substitute y = 1/5 x + 1 into x + 5y = -15

x + 5y = -15

x + 5(1/5 x + 1) = -15

x + x + 5 = -15

2x = -20

x = -10

Plug in x = -10 into y = 1/5 x + 1

y = 1/5 (-10) + 1

y = -2 + 1

y = -1

(-10 , -1)

Answer is the last one

(-10 , -1)

You might be interested in

The temperature is 5°C below zero. Which integer represents this temperature? −5 or +5

ahrayia [7]

-5 is the interference that represents this situation. Think of a time line, +5 would be 5 above 0.

3 0

3 years ago

Read 2 more answers

John wants to buy a skateboard that costs $150. The sales tax and 8.75%. What is the total cost of the equipment? $163.13 $153.1

kati45 [8]

Is any of the options that you put there.The correct answer would be $178.12

3 0

2 years ago

Suppose a random sample of 100 observations from a binomial population gives a value of pˆ = .63 and you wish to test the null h

irakobra [83]

Answer:

We conclude that the population proportion is equal to 0.70.

Step-by-step explanation:

We are given that a random sample of 100 observations from a binomial population gives a value of pˆ = 0.63 and you wish to test the null hypothesis that the population parameter p is equal to 0.70 against the alternative hypothesis that p is less than 0.70.

Let p = population proportion.

(1) The intuition tells us that the population parameter p may be less than 0.70 as the sample proportion comes out to be less than 0.70 and also the sample is large enough.

(2) So, Null Hypothesis, $H_0$ : p = 0.70 {means that the population proportion is equal to 0.70}

Alternate Hypothesis, $H_A$ : p < 0.70 {means that the population proportion is less than 0.70}

The test statistics that would be used here One-sample z-test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion = 0.63

n = sample of observations = 100

So, the test statistics = $\frac{0.63-0.70}{\sqrt{\frac{0.70(1-0.70)}{100} } }$

= -1.528

The value of z-test statistics is -1.528.

Now at 0.05 level of significance, the z table gives a critical value of -1.645 for the left-tailed test.

Since our test statistics is more than the critical value of z as -1.528 > -1.645, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the population proportion is equal to 0.70.

(c) The observed level of significance in part B is 0.05 on the basis of which we find our critical value of z.

4 0

3 years ago

The double number line shows that 4 pounds of tomatoes cost $14. How much does it cost for 1 pound of tomatoes

Anton [14]

Answer:

3.5

Step-by-step explanation:

You would do 14 divided by 4.

1pound =3.5

4 0

2 years ago

Read 2 more answers

Solve the system of equations using matrices. Use the Gauss- Jordan elimination method And find a solution set

KATRIN_1 [288]

$\begin{gathered} x+y+z=4 \\ x-y-z=0 \\ x-y+z=8 \\ \text{The system using matrix is} \\ \begin{bmatrix}{1} & {1} & {1}, & {4} \\ {1} & {-1} & {-1,} & {0} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix}\rightarrow F2=F2-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix} \\ \rightarrow F3=F3-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F2=-\frac{1}{2}F2 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F3=F3+2F2=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {2,} & {8}{}{}\end{bmatrix}\rightarrow F3=\frac{1}{2}F3 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F2=F2-F3=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \rightarrow F1=F1-F3=\begin{bmatrix}{1} & {1} & {0}, & {0} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F1=F1-F2 \\ =\begin{bmatrix}{1} & {0} & {0}, & {2} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \text{Therefore, the solution is }x=2,\text{ y=-2 and z=4} \end{gathered}$

6 0

1 year ago