All categories

2 years ago

$41.50 per day to rent a car and $ 0.10 for every mile driven. She plans to drive 225 miles, she has at most 230 to spend.

Mathematics

1 answer:

Sati [7]2 years ago

3 0

5 days to drive 225 miles
225 x .1= 22.50
230-22.50= 207.5
207.5/41.50= 5 days

You might be interested in

Urgent!!

Iteru [2.4K]

Answer:

82 is the median

Step-by-step explanation:

8 0

3 years ago

Find x, y, z, so that 41, 46, x, y, z has a mean of 50 and a mode of 45

nevsk [136]

Answer:

Step-by-step explanation:

as mode is 45 ,two of three is 45

i assume x =y=45

now (41+46+45+45+z)/5=50

177+z=50×5=250

z=250-177=73

so x=45

y=45

z=73

5 0

3 years ago

Read 2 more answers

A football team gained 5 yards and then last 7 yards. What real number represents the teams position relative to its original po

lana66690 [7]

Answer:

$\boxed{-2}}$

Step-by-step explanation:

Think of this situation as a number line in which the team starts at 0 yd.

Gaining yards is the positive direction and losing yards is negative.

Start at 0 yd and go 5 yd to the right. You are at +5.

Then go 7 yd left. You end up at -2.

+5 – 7 = -2

$\text{The real number that represents the team's position relative to its original position is $\boxed{\mathbf{-2}}$}$

5 0

3 years ago

At the movie theatre, child admission is $6.20 and adult admission is $9.30 . On Monday, 157 tickets were sold for a total sales

Andre45 [30]

C = child tickets
a= adult tickets

c + a = 157
6.20c + 9.30a = 1178.00

6.20c + 9.30a = 1178.00
-6.20c -6.20a = -973.40

3.10a = 204.60

a = 66 adult tickets

c + 66 = 157
-66. -66

c= 91 child tickets

8 0

3 years ago

An urn contains 8 red chips, 10 green chips, and 2 white chips. A chip is drawn and replaced, and then a second chip is drawn.

Harlamova29_29 [7]

Answer:

(A) 0.04

(B) 0.25

Step-by-step explanation:

Let R = drawing a red chips, G = drawing green chips and W = drawing white chips.

Given:

R = 8, G = 10 and W = 2.

Total number of chips = 8 + 10 + 2 = 20

$P(R) = \frac{8}{20}=\frac{2}{5}\\P(G)= \frac{10}{20}=\frac{1}{2}\\P(W)= \frac{2}{20}=\frac{1}{10}$

As the chips are replaced after drawing the probability of selecting the second chip is independent of the probability of selecting the first chip.

(A)

Compute the probability of selecting a white chip on the first and a red on the second as follows:

$P(1^{st}\ white\ chip, 2^{nd}\ red\ chip)=P(W)\times P(R)\\=\frac{1}{10}\times \frac{2}{5}\\ =\frac{1}{25} \\=0.04$

Thus, the probability of selecting a white chip on the first and a red on the second is 0.04.

(B)

Compute the probability of selecting 2 green chips:

$P(2\ Green\ chips)=P(G)\times P(G)\\=\frac{1}{2} \times\frac{1}{2}\\ =\frac{1}{4}\\ =0.25$

Thus, the probability of selecting 2 green chips is 0.25.

(C)

Compute the conditional probability of selecting a red chip given the first chip drawn was white as follows:

$P(2^{nd}\ red\ chip|1^{st}\ white\ chip)=\frac{P(2^{nd}\ red\ chip\ \cap 1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\=\frac{P(2^{nd}\ red\ chip)P(1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\= P(R)\\=\frac{2}{5}\\=0.40$

Thus, the probability of selecting a red chip given the first chip drawn was white is 0.40.

6 0

3 years ago