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3 years ago

Help please? Brainliest + points

Mathematics

2 answers:

ryzh [129]3 years ago

5 0

Hi!

<h3>Let's go through each answer and see if it's correct. </h3>

A. Taxi A is always the better deal.

This is false, because at the beginning Taxi B is the better deal.

B. Taxi B is always the better deal.

This is false, because at the end, taxi A is the better deal.

C. Taxi A is the better deal for the first 7.5 miles.

This is false, because if you look at the graph, you'll see that Taxi B is the better deal for the first 7.5 miles.

D. Taxi B is the better deal for the first 7.5 miles.

<h2>The answer is D. Taxi B is the better deal for the first 7.5 miles. </h2>

Hope this helps! :)

-Peredhel

expeople1 [14]3 years ago

4 0

D because is costs less until it gets to 7.5 on the x axis

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Combine like terms to create an equivalent expression of 3.26d + 9.75d - 2.65

sleet_krkn [62]

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3 years ago

A certain type of automobile battery is known to last an average of 1140 days with a standard deviation of 80 days. If 400 of th

Sever21 [200]

Answer:

Mean = $\mu = 1140$

Standard deviation = $\sigma = 80$

Find the probabilities for the average length of life of the selected batteries.

A)The average is between 1128 and 1140.

We are supposed to fidn P(1128<x<1140)

Formula : $Z=\frac{x-\mu}{\sigma}$

At x = 1128

$Z=\frac{1128-1140}{80}$

$Z=-0.15$

Refer the z table for p value

P(x<1128)=0.4404

At x = 1140

$Z=\frac{1140-1140}{80}$

$Z=0$

Refer the z table for p value

P(x<1140)=0.5

So,P(1128<x<1140)=P(x<1140)-P(x<1128)=0.5-0.4404=0.0596

Hence the probabilities for the average length of life of the selected batteries is between 1128 and 1140 is 0.0596

B)The average is greater than 1152.

P(x>1152)

At x = 1128

$Z=\frac{1152-1140}{80}$

$Z=0.15$

Refer the z table for p value

P(x<1152)=0.5596

So,P(x>1152)=1-P(x<1152)=1-0.5596=0.4404

Hence the probabilities for the average length of life of the selected batteries is greater than 1152 is 0.4404

C) The average is less than 940.

P(x<940)

At x = 940

$Z=\frac{940-1140}{80}$

$Z=-2.5$

Refer the z table for p value

P(x<940)=0.5596

Hence the probabilities for the average length of life of the selected batteries is less than 940 is 0.5596

7 0

3 years ago

Nelly spends 25% of her free times surfing the internet. If she spends approximately 1 hour per day surfing the internet,how man

Alex Ar [27]

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4 years ago