A farmer has 212 acres of land for sale. He divides the land into 8 equal sections. About how many acres are in each section?

HHHHHHHHHEEEEEELLLLLLLLPPPPPPPP how do I do this

lyudmila [28]

Answer:

1/6

Step-by-step explanation:

2/3 times 1/4 is 2/12. 2/12 simplified is 1/6, so 1/6 is your answer.

7 0

3 years ago

The school band collected $2,892 for 480 tickets sold to their spring concert. Adult tickets were $8 each and student tickets we

kakasveta [241]

A + s = 480......s = 480 - a
8a + 5s = 2892

8a + 5(480 - a) = 2892
8a + 2400 - 5a = 2892
8a - 5a = 2892 - 2400
3a = 492
a = 492/3
a = 164 <=== there were 164 adult tickets sold

5 0

4 years ago

Read 2 more answers

Tom works in the cold food department in a supermarket. The thermometer shows the temperature of one of the freezers. The freeze

serg [7]

To return back to -18°C from -10°C, you would have to lower the temperature by -8°C to compensate for the difference.

Hope that helps!

5 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

Find the measure of <4. { 44 158° { 44 = [?]

VLD [36.1K]

Answer:

<4 = 22

Step-by-step explanation:

<4 = x

x + 158 = 180

x = 22

4 0

3 years ago