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AURORKA [14]
2 years ago
11

Natasha had a $922.93 balance on her credit card at the beginning of September. Her credit card has an APR of 9.89%, compounded

monthly, and a minimum monthly payment of 3.08% of the total balance. The following table shows Natasha’s credit card purchases over the next two months.
Mathematics
1 answer:
avanturin [10]2 years ago
5 0

Answer:

C

Step-by-step explanation:

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The Office of Student Services at a large western state university maintains information on the study habits of its full-time st
vredina [299]

Answer:

a. 0.50

Step-by-step explanation:

The standard error of the mean is the standard deviation of the population divided by the square root of the sample size.

In this problem, we have that:

Standard deviation of the population: 6 hours

Sample size: 144

Square root of 144 is 12.

So the standard error of the sample mean is 6/12 = 0.5.

4 0
1 year ago
A home’s value increases at an average rate of 5.5% each year. The current value is $120,000. What function can be used to find
Wewaii [24]

Answer:

y=120,000(1.055)^x

Step-by-step explanation:

we know that

The equation of a exponential growth function is given by

y=a(1+r)^x

where

y is the value of the home

x is the number of years

a is the initial value

r is the rate of change

we have

a=\$120,000\\r=5.5\%=5.5/100=0.055

substitute

y=120,000(1+0.055)^x

y=120,000(1.055)^x

7 0
1 year ago
PLZ HELP ME ASAP The box plot shows the typing speed (in words per minute without errors) of the contestants in a typing contest
Paraphin [41]
1. 95-87 = 8 /2 = 4

2. 91 - 82 = 9 /2 = 4.5 =5

3. 91 - 87 = 4

4. B 1 times
6 0
1 year ago
Read 2 more answers
Cameron wondered if the average score on a final exam was different between those who texted on a regular basis during the lectu
Margarita [4]

Answer:

c. A two-tailed test should be performed since the alternative hypothesis states that the parameter is not equal to the hypothesized value.

Step-by-step explanation:

Let p1 be the average score on a final exam who texted on a regular basis during the lectures for a particular class

And p2 be the average score on a final exam who did not texted at all during the lectures for a particular class

According to the Cameron's point of interest, null and alternative hypotheses are:

H_{0}: p1 = p2

H_{a}: p1 ≠ p2

Two tailed test should be performed since the alternative hypothesis states that the parameter is not equal to the hypothesized value.

3 0
2 years ago
You are traveling down a country road at a rate of 95 feet/sec when you see a large cow 300 feet in front of you and directly in
emmainna [20.7K]

Answer:

1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.

2) See attached picture.

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in ft/s^{2}

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) j(t)=\left \{ {{95t-9t^{2}; 0\le t

j'(t)=\left \{ {{95-18t; 0\leq t

(see attached picture for graph)

Step-by-step explanation:

1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:

j(t)=95t-9t^2

j'(t)=95-18t

and we set the first derivative equal to zero so we get:

95-18t=0

and solve for t

-18t=-95

t=\frac{95}{18}

t=5.28s

so now we calculate the position of the car after 5.28 seconds, so we get:

j(5.28)=95(5.28)-9(5.28)^{2}

j(5.28)=250.69ft

so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.

2) For answer 2 I take the second derivative of the function so I get:

j(t)=95t-9t^{2}

j'(t)=95-18t

j"(t)=-18

and then we graph them. (See attached picture)

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in ft/s^{2}

3)  yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) j(t)=\left \{ {{95t-9t^{2}; 0\le t

j'(t)=\left \{ {{95-18t; 0\leq t

(see attached picture for graph)

5 0
1 year ago
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