Please help me!!! :(

Rick rents a car from a company that charges him $25 per day plus 10 cents per mile driven. Which expression can represent the a

Lemur [1.5K]

C=25x+.10
C is the total cost. To get the total cost you add the consistent fee of $25 per day, times the number of days. You add $.10 for every mile driven.

4 0

2 years ago

8. Marvin's son is 3 ft tall. He casts a shadow of 1 foot. Marvin cast a

taurus [48]

Answer:

C. 6 ft

Step-by-step explanation:

If Marvin's son is 3 feet tall with a 1 foot shadow, Marvin would be 6 feet tall with a 2 foot shadow.

7 0

3 years ago

Read 2 more answers

Laura jog around the soccer field that 80m by 120m.how many yards does she goes around the field two times?

exis [7]

A rectangle has two couples of equal sides, so a soccer fields has two 80-meters sides and two 120-meters sides.

This leads to a perimeter of

$120+120+80+80=400 \text{ m}$

If you go around it twice, you'll run 800 meters

8 0

3 years ago

The proportion of defective computers built by Byte Computer Corporation is 0.15. In an attempt to lower the defective rate, the

FinnZ [79.3K]

Answer:

a) Null hypothesis : H₀: the proportion of defective item of computer has been lowered. That is P < 0.15

Alternative hypothesis: H₁: The proportion of defective item of computer

has been higher. That is P> 0.15 (Right tailed test)

b) Test statistic $Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }$

c) Calculate the value of the test statistic = 0.991

d) The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57

e) Null hypothesis accepted at 0.01 level of significance

f) we accepted null hypothesis.

Hence the proportion of defective item of computer has been lowered.

Step-by-step explanation:

Step(i):-

Given the sample size 'n' = 42

Given random sample of 42 computers were tested revealing a total of 4 defective computers.

The defective computers 'x' = 4

The sample proportion of defective computers

$p = \frac{x}{n} = \frac{4}{42} = 0.095$

Given The Population proportion 'P' = 0.15

The level of significance ∝=0.01

Step(ii):-

a) Null hypothesis : H₀: the proportion of defective item of computer has been lowered. That is P < 0.15

Alternative hypothesis: H₁: The proportion of defective item of computer

has been higher. That is P> 0.15 (Right tailed test)

b)

Test statistic $Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }$

c)

$Z = \frac{0.095-0.15}{\sqrt{\frac{0.15(0.85)}{42} } }$

$z = \frac{-0.055}{\sqrt{0.00303} } = - 0.9991$

Calculate the value of the test statistic Z = - 0.9991

|Z| = |- 0.9991| = 0.991

Step(iii):-

d)

The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57

e) Calculate the value of the test statistic Z = 0.991 < 2.57 at 0.01 level of significance.

Conclusion:-

Hence the null hypothesis is accepted at 0.01 level of significance.

f)

The proportion of defective item of computer has been lowered.

6 0

3 years ago

kelly wants to build a pacman shaped pool in her back yard pictured to the right (3/4 of a circle). she needs to have a gate aro

Andreas93 [3]

You need to find the circumference of the whole circular pool and then subtract from it the arc length that is 1/4 of the circumference. The circumference formula is C = (3.14)(d), which in our case is 3.14(40) which is 125.6. Now we need the arc length we need to take away. 1/4 of a circle corresponds to a 90 degree angle, so in the formula for arc length, we have this: $\frac{90}{360} (3.14)(40)$ which equals 31.4. Now subtract that from the circumference of the whole circle and you get the outside of the circle minus that arc length. That's 94.2. But we have to go in the radius times 2 to enclose the pie shaped piece we cut away. So we have 134.2. She should've just stuck with enclosing the circle; she would have used less fencing!

4 0

2 years ago