The Numbers On This Ruler Represent Centimeters

Five less than one fourth of my age is 12

Ber [7]

5 less than (-5) of 1/4 of my age (age/4) is 12 (=12)

-5+a/4=12
add 5 to both sides
a/4=17
times 4 both isdes
a=68

your age is 68

6 0

2 years ago

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Please help now i need this quick

algol [13]

Answer:

5= b 6=d

Step-by-step explanation:

7 0

3 years ago

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Ill give a brainliest i need help its due today

aliya0001 [1]

Answer:

w=15

Step-by-step explanation:

-24=-3w+21

We move all terms to the left:

-24-(-3w+21)=0

We get rid of parentheses

3w-21-24=0

We add all the numbers together, and all the variables

3w-45=0

We move all terms containing w to the left, all other terms to the right

3w=45

w=45/3

w=15

8 0

3 years ago

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Find the volume of the prism ASAP

tino4ka555 [31]

The answer is 60 you just multiply the length and width and height

5 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

2 years ago