If f(1)=160 and f(n+1)=2(n) what is f(4)

jamal is going to borrow 14000 from his credit union to buy a used car. the APR is 7.0% and the length of the loan is 4 years. h

ELEN [110]

The formula to use here would be:

A = P(1+r)^t where P is the principal (starting) amount, r is the rate of interest, and t is time. Knowing this, plug in your values:

A = 14,000(1+0.07)^4 once worked out, you should get $18,351.14414 < this is the total amount he would be paying at the end of 4 years. to find the interest, subtract this amount from the principal amount (14,000):

$18,5351.14414 - 14,000 = 4,351.14414 < this is the amount of interest. Round your answer.

Answer: Jamal will pain $4,351.14 interest.

8 0

3 years ago

HELPLPLPPLPPLPPLPPLPLPLPL

kipiarov [429]

Answer:

Bro the answer is LITERALLY 1

Step-by-step explanation:

4 0

3 years ago

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Beth wants to buy a shirt that costs $40. She has a 20% discount. The sales tax is 8%.

Aleks04 [339]

Answer:

$34.56

Step-by-step explanation:

She will save 8 dollars from the discount.

The tax will be $0.56.

The final price of the shirt will be $34.56

6 0

3 years ago

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Ms.Rollins had and end of the year pizza party for the chess team.At the party,every 2 students had 5 slices of pizza

anyanavicka [17]

Answer:

Step-by-step explanation:

What's your question

6 0

3 years ago

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You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confide

salantis [7]

Answer:

With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

Step-by-step explanation:

We are given that a random sample of 60 home theater systems has a mean price of$131.00. Assume the population standard deviation is$18.80.

Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean price = $131

$\sigma$ = population standard deviation = $18.80

n = sample of home theater = 60

$\mu$ = population mean

Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $131-1.645 \times {\frac{18.8}{\sqrt{60} } }$ , $131+1.645 \times {\frac{18.8}{\sqrt{60} } }$ ]

= [127.01 , 134.99]

Therefore, 90% confidence interval for the population mean is [127.01 , 134.99].

Now, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean price = $131

$\sigma$ = population standard deviation = $18.80

n = sample of home theater = 60

$\mu$ = population mean

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $131-1.96 \times {\frac{18.8}{\sqrt{60} } }$ , $131+1.96 \times {\frac{18.8}{\sqrt{60} } }$ ]

= [126.24 , 135.76]

Therefore, 95% confidence interval for the population mean is [126.24 , 135.76].

Now, with 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

7 0

3 years ago