Please explain your answer as well. Thx!!!

Katie walked a total of 68 kilometers by making 34 trips to school. How many trips will Katie have to make in all to walk a tota

ZanzabumX [31]

Find the distance of each trip:

68 km / 34 trips = 2 km per trip

Divide total distance by distance per trip:

348 km / 2 km per trip = 174 trips

Answer: 174 trips

5 0

3 years ago

Please help! Thanks in advance.

alukav5142 [94]

In this problem, we need to plug in the given x values for $f(x)=0$ and find a and b.

When we plug in 1, we get: $2 \times {1}^{4} - 5 \times {1}^{3} - 14 \times {1}^{2} + a \times 1 + b = 0$

Simplify:
$2 - 5 - 14 + a + b = 0$
$- 17 + a + b = 0$
$a + b = 17$

We got our first statement about the values of the variables. If we find one more we can find those 2 variables.

We have another given root: 4.

Plug it in:
$2 \times {4}^{4} - 5 \times {4}^{3} - 14 \times {4}^{2} + a \times 4 + b = 0$
$512 - 320 - 224 + 4a + b = 0$
$- 32 + 4a + b = 0$
$4a + b = 32$

Now we have our second one. We can combine them:
$a + b = 17 \\ 4a + b = 32$

I use elimination method which is easier here.

Multiply the top equation by -1: $- a - b = - 17 \\ 4a + b = 32$

Add them up:
$3a = 15$

Simplify:
$a = 5$

Now we have a, we can plug in one of those equations to find b:
$5 + b = 17$
$b = 17 - 5$
$b = 12$

So, the answers are $a=5$ and $b=12$ .

6 0

4 years ago

Problem is in the attached picture. Thanks in advance!

bogdanovich [222]

Answer:

Width= 120 ft

Length= 285 ft

Step-by-step explanation:

W= width and (3w-75)= length since it's 75 less than 3 times the width

w+w+(3w-75)+(3w-75)=810

8w-150=810

8w=960

w=120

So the width is 120 ft

Then substitute the 120 in for w in the length

L= 3(120)-75

L=360-75

L=285

So the length is 285 ft

You can check because 120+120+285+285=810, and because 120×3-75=285!

6 0

3 years ago

If one gallon of paint covers 350 square feet how many gallons will you need to cover a wall that is 138 inches by 102 inches

frosja888 [35]

138in x 102in = 14076in sq.
14076/350=40.217
41 gallons of paint.

6 0

4 years ago

A company claims that the mean weight per apple they ship is 120 grams with a standard deviation of 12 grams. Data generated fro

Veronika [31]

Answer:

There is no sufficient evidence to reject the company's claim at the significance level of 0.05

Step-by-step explanation:

Let $\mu$ be the true mean weight per apple the company ship. We want to test the next hypothesis

$H_{0}:\mu=120$ vs $H_{1}:\mu\neq 120$ (two-tailed test).

Because we have a large sample of size n = 49 apples randomly selected from a shipment, the test statistic is given by

$Z=\frac{\bar{X}-120}{\sigma/\sqrt{n}}$ which is normally distributed. The observed value is

$z_{0}=\frac{122.5-120}{12/\sqrt{49}}=1.4583$ . The rejection region for $\alpha = 0.05$ is given by RR = {z| z < -1.96 or z > 1.96} where the area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well. Because the observed value 1.4583 does not fall inside the rejection region RR, we fail to reject the null hypothesis.

5 0

3 years ago