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4 years ago

10

Pat is 19 years old. He buys 25/50/50 liability insurance, collision insurance with a $250 deductible, and comprehensive insuran

ce with a $100 deductible. What is his total annual premium?

1 answer:

beks73 [17]4 years ago

5 0

Answer:

$2337

Step-by-step explanation:

this is the answer

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If a polynomial function f(x) has roots 3+Sqrt(5) and -6, what must be a factor of f(x)

neonofarm [45]

Answer:

The two factors of the given polynomial f(x) are (x - 3 -√5) and (x +6) .

Step-by-step explanation:

Here, the roots of the given polynomial f(x) are

Root 1 = 3 + √5

and Root 2 = (-6)

Now,as we know that if x = a is the root of the any given polynomial p(x), then the expression (x-a) is the FACTOR of the polynomial p(x).

Now, here root 1 : x = 3 + √5

So, x - (3 + √5 ) is the FACTOR of the f(x)

or, ( x - 3 -√5) is the first factor of f(x).

Also, here root 2 : x = (-6)

So, x - (-6 ) = x + 6 is the FACTOR of the f(x)

or, ( x +6 ) is the second factor of f(x).

Hence, the two factors of the given polynomial f(x)

are ( x - 3 -√5) and ( x +6 ) .

5 0

4 years ago

A car is 170 inches a tuck is 8% longer how long is the truck

atroni [7]

170+8%= 183.6 inches
hope this helped

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3 years ago

A random sample of 60 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact te

pentagon [3]

Answer:

The 95% confidence interval on the true proportion of helmets of this type that would show damage from this test is (0.169, 0.397).

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 60, \pi = \frac{17}{60} = 0.283$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.283 - 1.96\sqrt{\frac{0.283*0.717}{60}} = 0.169$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.283 + 1.96\sqrt{\frac{0.283*0.717}{60}} = 0.397$

The 95% confidence interval on the true proportion of helmets of this type that would show damage from this test is (0.169, 0.397).

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3 years ago

Will mark brainliest!! please help me

lions [1.4K]

The correct answer is A !

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3 years ago

What is the minimum value of the graph of y = cos(x)?

Nostrana [21]

Answer:-1

Step-by-step explanation:

C

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3 years ago