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3 years ago

Solve -4r + 5=25 for the value of r

Mathematics

2 answers:

alexdok [17]3 years ago

8 0

-4r+5=25 -4r=20 r=-5
typing this to get to 20 characterssss

Vedmedyk [2.9K]3 years ago

5 0

First you have to set r by itself and to do this you have to subtract 5 from both sides (because 5 is positve and in order to get rid of a positive is with a negative) now we have -4r=20, now divide both sides by -4 to leave r by itself, when you do you will end up with r=-5

Hope this helps

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A consumer organization estimates that over a 1-year period 20% of cars will need to be repaired once, 5% will need repairs

gtnhenbr [62]

Answer:

a) 0.5476

b) 0.0676

c) 0.4524

Step-by-step explanation:

Given this information, we can conclude that 74% of the cars won't need any repairs over a 1-year period (100 - 20 - 5 - 1 = 74%). And 26% will need at least 1 repair over a 1-year period.

P(car doesn't need repair) = 0.74

P (car needs repair) = 0.26

If you own two cars, the probability that:

a) Neither will need repair:

We need that car 1 won't need repair AND car 2 won't need repair.

=P(Car 1 doesn't need repair) x P(Car 2 doesn't need repair)

= 0.74 x 0.74 = 0.5476

The probability that neither will need repair is 0.5476.

b) Both will need repair:

We need that car 1 needs repair AND car 2 needs repair.

P(Car 1 needs repair) x P(Car 2 needs repair)

= 0.26 x 0.26 = 0.0676

The probability that both will need repair is 0.0676

c) At least one car will need repair

Car 1 needs repair or Car 2 needs repair or both need repair.

To solve this one, it's easier to use the complement of P(neither needs repair)

1 - P(neither needs repair)

1 - (0.74)(0.74) = 1 - 0.5476 = 0.4524

The probability that at least one car will need repair is 0.4524

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3 years ago

US average math SAT scores follow a normal distribution with a mean of 505 and a standard deviation of 112. A sample of 64 enter

Hitman42 [59]

Answer:

The claim that the scores of UT students are less than the US average is wrong

Step-by-step explanation:

Given : Sample size = 64

Standard deviation = 112

Mean = 505

Average score = 477

To Find : Test the claim that the scores of UT students are less than the US average at the 0.05 level of significance.

Solution:

Sample size = 64

n > 30

So we will use z test

$H_0:\mu \geq 477\\H_a:\mu < 477$

Formula : $z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

$z=\frac{505-477}{\frac{112}{\sqrt{64}}}$

$z=2$

Refer the z table for p value

p value = 0.9772

α=0.05

p value > α

So, we accept the null hypothesis

Hence The claim that the scores of UT students are less than the US average is wrong

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3 years ago