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3 years ago

3. The amount was $120 at 5% rate for 4 years.

Mathematics

1 answer:

Inga [223]3 years ago

5 0

Answer:

Go to photo math thank me later

Step-by-step explanation:

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Niu has decorated x cards. He started with 24 stickers, and he used 4 stickers per card. Which expressions can we use to describ

Kisachek [45]

Answer:

$24-4x$

Step-by-step explanation:

We have been given that Niu has decorated x cards. He started with 24 stickers and he used 4 stickers per card. So stickers used in x cards will be 4x.

As initial number of stickers was 24, so we can find number of stickers left after decorating x cards by subtracting number of stickers used in x cards from 24.

$\text{Number of stickers that Niu has left}=24-4x$

Therefore, the expression $24-4x$ will represent the number of stickers that Niu has left after decorating x cards.

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4 years ago

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A number sentence is shown below: 45 1/3 / 1/6 272 Part A: Create a story or context for this number sentence. (4 points) Part B

tino4ka555 [31]

Answer:

1/3 / 1/6 272 Part A: Create a story or context for this number sentence.

Step-by-step explanation:

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3 years ago

Answer please !!! i need help asap

maksim [4K]

Answer:

A. Undefined will be your answer

Step-by-step explanation:

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3 years ago

Read 2 more answers

Whats the answer for this 4b²+20b+25

WITCHER [35]

4b²+20b+25=0

Divide everything by 4

b²+5b+ 6.25=0

use the quadratic formula
x= (-b+ or - √b²-4ac) /2a
x= (-5 + or - √5²-4*1*6.25) /2(1)
x= -5/2 so
in this case, b= -5/2

7 0

3 years ago

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