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2 years ago

Students on a field trip at an amusement park were asked their grade in school. The table shows the results of the survey.

Mathematics

2 answers:

muminat2 years ago

6 0

C) I KNOW IT IS!!
SO DO THIS ONE!!

stira [4]2 years ago

3 0

Answer: Option 'C' is correct.

Step-by-step explanation:

Since we have given that

Grade Number of Students

4th 45

5th 25

6th 60

7th 50

8th 60

9th 45

Total number of students is given by

45+25+60+50+60+45 = 285

So, the central angle for 9th grade would be

$\dfrac{45}{285}\times 360^\circ\\\\=56.8^\circ\\\\\approx 57^\circ$

Hence, Option 'C' is correct.

You might be interested in

Find the limit of the function by using direct substitution. limit as x approaches four of quantity x squared plus three x minus

Yuliya22 [10]

$\lim_{x \to 4} f(x)=x^2+3x-1$
just subsitute
f(4)=4²+3(4)-1
f(4)=16+12-1
f(4)=28-1
f(4)=27

it approaches 27 as x approaches 4

5 0

3 years ago

Read 2 more answers

Jocelyn’s plane travelled 500 mph for 7.3 hours. How far did her plane travel?

kogti [31]

Answer:

Jocelyn's plane travelled 3650 miles.

Step-by-step explanation:

mph stands for miles per hour.

Let's set up an equation.

total miles = speed(hours)

Jocelyn's plane travelled 500 mph for 7.3 hours.

total miles = 500(7.3)

Multiply.

total miles = 3650

Jocelyn's plane travelled 3650 miles.

Hope this helps!

3 0

3 years ago

Please helpppp:////!!!

Flauer [41]

I believe the answer is C>=1,499.

3 0

3 years ago

Read 2 more answers

Your college roommate receives a pay raise at her part-time job from $9 to $11 per hour. She used to work 25 hours per week, but

Fynjy0 [20]

Answer:

d. backward sloping.

Step-by-step explanation:

The roommate receives a pay raise at her part-time job from $9 to $11 per hour and now decides to work 20 hours per week instead to 25 hours per week.

We can see that her rate is decreasing, so her labor supply curve is backward sloping or also called backward bending.

It is mostly found that worker's labor supply curve, slopes upward when they get lower wages and the curve bends backward when they get higher wages because a worker tends to work less when his wage rate rises.

6 0

3 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$