Write an equation in slope intercept form with a slope of 9/10 and a y-intercept of 3.

Alex is writing statements to prove that the sum of the measures of interior angles of triangle PQR is equal to 180°. Line m is

Crank

Answer: The stuff in pink is the stuff I found based on the information given. The pink 40 comes from the blue 40. The reason that both of those angles are congruent is because they are alternate interior angles. This is the same reason about the angles whose measurements are 60. So the following things are true that we found that weren't already given: Angle PQR is 40 Angle PRQ is 60

Step-by-step explanation:

8 0

3 years ago

A rental car costs $50 for one day plus an additional $0.25 per mile. What is the cost of renting a car for one day and driving

charle [14.2K]

The equation for this problem would be 50 + .25x, so if x is equal to 75, then the total would be $68.75

8 0

4 years ago

Read 2 more answers

I tried two ways to figure out 206-48 but both answers were different

konstantin123 [22]

158 is the answer to your question

7 0

4 years ago

Read 2 more answers

The distance d (in miles) you travel in a car is given by the two equations shown, where t is the time (in hours) and g is the n

lana [24]

We have two equations d=55t and d=20g

The left sides of this equations are the same than the right sides are equal too

a) 55t=20g

b) g=(55/20)t

c) t=6h => g=(55/20) *6 = 165/10=16.5 gallons

d= 55t= 55*6=330 miles or d=20g=20*16.5=330 miles

Good luck!!!

4 0

3 years ago

Read 2 more answers

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