What is the value of EBD?

Suppose you deposit $1,200 into a bank account. The account earns yearly simple interest at a rate of 1 3/4%. How many years wil

igor_vitrenko [27]

Answer:

6 years

Step-by-step explanation:

1.75% of 1,200 = 21

so 1 year= 21

21+21= 42 2 years

42+21=63 3 years

63+21=84 4 years

81+21=105 5 years

105+21=126 6 years

Or you can do

21x6=126 so it has to equal 6 years

4 0

3 years ago

HELP ME PLEASEEEEEEE

igomit [66]

Answer:a

Step-by-step explanation:The temperature rose by 80 so up the number line and dropped 20 so down the number line to 60

4 0

3 years ago

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a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was

Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

Let $\bar X$ = sample average time spent studying

The z-score probability distribution for sample mean is given by;

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

where, $\mu$ = population mean hours spent studying = 25 hours

$\sigma$ = standard deviation = 15 hours

n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < $\bar X$ < 30 hours)

P(29 hours < $\bar X$ < 30 hours) = P( $\bar X$ < 30 hours) - P( $\bar X$ $\leq$ 29 hours)

P( $\bar X$ < 30 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z < 2) = 0.97725

P( $\bar X$ $\leq$ 29 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z $\leq$ 1.60) = 0.94520

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.

Therefore, P(29 hours < $\bar X$ < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0

3 years ago

A ball is launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall. The equation for the ball’

xz_007 [3.2K]

When the ball is in the ground, its height is basically equal to zero thus making our equation,

-16t^2 + 272t + 1344 = 0

Simplifying the equation will give us,

- t^2 + 17t + 84 = 0 or t^2 – 17t – 84 = 0

Factoring out the equation will give us,

(t – 21)(t + 4) = 0

Thus, t = 21 or t = -4. -4 is an extraneous root. Thus, the answer is t = 21.

Answer: 21 seconds

5 0

3 years ago

What is 6 1/8 minus 2 1/12

Bezzdna [24]

Answer:

4 1/24

Step-by-step explanation:

4 0

3 years ago

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