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Sophie [7]
3 years ago
6

(20 points, please help)

Mathematics
2 answers:
jeka943 years ago
8 0
I think it’s c
I know the answer


kvv77 [185]3 years ago
6 0
I think that the answer is C
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A manufacturer of pickup trucks is required to recall all the trucks manufactured in a given year for the repair of possible def
Crank

Answer: 2%

Step-by-step explanation:

Let A be the event of having defective steering and B be the vent of having defective brake linings.

Given: P(A)  = 0.03  P(B) = 0.05

P(neither A nor B ) = 0.94

Using formula: P(either A nor B) = 1- P(neither A nor B )

= 1-0.94

i.e.  P(either A nor B) =0.06

Using formula:P(A and B) = P(A)+P(B)-P(either A or B)

P(A and B) =0.03+0.05-0.06

= 0.02

Hence, the percentage of the trucks have both defects = 2%

8 0
3 years ago
What is the value of 8 in 281
enot [183]
The value is 80 because 8 is in the tens place.
4 0
3 years ago
Read 2 more answers
a rain gutter is 24 feet long, 4 inches in height, 3 inches at base and 6 inches at top. how many gallons of water will it hold?
Lubov Fominskaja [6]

Answer:1.8701

Step-by-step explanation:

Area (A) of trapezium as marked is =\frac{1}{2}\left ( sum\ of\ parallel\ sides\right )\times height

A=\frac{1}{2}\left ( 3+6\right )\times 4

A=18in^2

Now for Volume we have to multiply by length because area is same across the length.

volume(v)=A\times length

Volume(v)=18\times 24

Volume(v)=432 in^3

Volume(v)=1.8701 gallon

3 0
3 years ago
He exponential model describes the​ population, A, of a country in​ millions, t years after 2003. Use the model to determine the
Tasya [4]

Complete question :

The exponential model A=16.2e^0.01t describes the population, A, of a country in millions, t years after 2003. What was the population in 2003?

Answer:

16.2 million

Step-by-step explanation:

Given the equation :

A=16.2e^0.01^t

Where, t = number of years after 2003

The population in year 2003 ; can be obtained thus ;

t = 2003 - 2003 = 0

Put t = 0 in the equation :

A(0) =16.2e^0.01^0

A(0) = 16.2 * 1

A(0) = 16.2

Hence, population in 2003 is 16.2 million

7 0
3 years ago
PLEASE HELP!
Aleks04 [339]

Answer:

Option c:

0.25 \pm 1.645\sqrt{\frac{0.25*(1-0.25)}{90}}

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

For this problem, we have that:

You have access to first year enrolment records and you decide to randomly sample 119 of those records. You find that 89 of those sampled went on to complete their degree. This means that n = 119, \pi = \frac{89}{119} = 0.7478.

90% confidence level

So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

Sample of 90 people with a sample proportion of 0.25

This means that n = 90, p = 0.25.

Confidence interval:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

0.25 \pm 1.645\sqrt{\frac{0.25*(1-0.25)}{90}}

Which is option c.

6 0
3 years ago
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