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3 years ago

Pls Help!!!!!

Mathematics

2 answers:

lisabon 2012 [21]3 years ago

6 0

Answer:

That is good! Just plug them in the graph and you should be good. Also this should be a parabola correct?

Step-by-step explanation:

mestny [16]3 years ago

4 0

Answer: (-10, 23)

Step-by-step explanation:

You might be interested in

6 out of 10 students completed their homework. What is the ratio of students who did not do their homework to those who did?

ludmilkaskok [199]

Answer:

2/3 or 2:3

Step-by-step explanation:

If 6 out of 10 students completed the homework then 4 out of 10 students did not complete the homework.

So the answer is 4/6 but we are not done yet.

The 4 comes from people who didn't do it and the 6 comes from people who did.

Since its asking the ratio of students who did not do their homework to those who did the 4 comes first.

As I said we are not done yet since we have to simplify.

4/6=2/3

So the answer is 2/3 or 2:3

5 0

4 years ago

Robin Hartman earns $600 per week plus 3% of sales over $6,500. Robin's week sales are $14,000. How much does Robin earn

Karolina [17]

Answer:

$825

Step-by-step explanation:

Given that :

Weekly earning = $600 plus 3% of sales over $6500

If weekly sales = $14000

How much will Robin earn

Robin's earning will be:

Flat weekly pay + 3% of amount over 6500

$600 + 3%(weekly sales - 6500)

$600 + 0.03(14000 - 6500)

$600 + 0.03(7500)

$600 + $225

= $825

5 0

3 years ago

0.0943 0.9403 0.9043 least to greatest

BartSMP [9]

0.0943 then .9043 then .9403. Hope that helps

6 0

3 years ago

Read 2 more answers

Which algebraic expression has a term with a coefficient of 3?

Mariulka [41]

Answer:

D. 3y+1

Step-by-step explanation:

The coefficient is the number in front of the variable

3y has a coefficient of 3

3 ( y-6) has a factor of 3 since it is multiplied by the other term ( y-6)

4 0

3 years ago

Randomly selected 110 student cars have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly select

monitta

Answer:

1. Yes, there is sufficient evidence to support the claim that student cars are older than faculty cars.

2. The 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

Step-by-step explanation:

We are given that randomly selected 110 student cars to have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly selected 75 faculty cars to have ages with a mean of 5.3 years and a standard deviation of 3.7 years.

Let $\mu_1$ = mean age of student cars.

$\mu_2$ = mean age of faculty cars.

So, Null Hypothesis, $H_0$ : $\mu_1 \leq \mu_2$ {means that the student cars are younger than or equal to faculty cars}

Alternate Hypothesis, $H_A$ : $\mu_1>\mu_2$ {means that the student cars are older than faculty cars}

(1) The test statistics that will be used here is Two-sample t-test statistics because we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$ = sample mean age of faculty cars = 5.3 years

$s_1$ = sample standard deviation of student cars = 3.6 years

$s_2$ = sample standard deviation of student cars = 3.7 years

$n_1$ = sample of student cars = 110

$n_2$ = sample of faculty cars = 75

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(110-1)\times 3.6^{2}+(75-1)\times 3.7^{2} }{110+75-2} }$ = 3.641

So, the test statistics = $\frac{(8-5.3)-(0)} {3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} } }$ ~ $t_1_8_3$

= 4.952

The value of t-test statistics is 4.952.

Since the value of our test statistics is more than the critical value of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we support the claim that student cars are older than faculty cars.

(2) The 98% confidence interval for the difference between the two population means ( $\mu_1-\mu_2$ ) is given by;

98% C.I. for ( $\mu_1-\mu_2$ ) = $(\bar X_1-\bar X_2) \pm (t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} })$

= $(8-5.3) \pm (2.326 \times 3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} })$

= $[2.7 \pm 1.268]$

= [1.432, 3.968]

Here, the critical value of t at a 1% level of significance is 2.326.

Hence, the 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

7 0

3 years ago