There are 49 students had their lunch in the cafeteria. One-

Please help 100 points

muminat

Answer:

Type that out please I can't read it sideways

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

If (2) - 2x - 6, which of these is the inverse of x)?

Romashka [77]

Answer:

Step-by-step explanation:

y=-2x+6.

since it is inverse the gradient has to be the opposite of the base equation : y =2x-6.

dont forget to put it as the brainliest answer if it helps.

7 0

3 years ago

A barrel of tomato sauce has spilled on a tile floor. The sauce flow can be expressed with the function r(t) = 2t, where t repre

IrinaVladis [17]

For the answer to the question above, I'll show my solution for the answer below.
Area A = pi r^2
The rate of change of area = d A/ d t = 2 pi r d r/d t 
r= 2t, d r = 2 d t 
This gives 
d A/ d t = 2 pi 2t 2 d t = 8 pi t dt 
Integrate it 
A= 4 pi t^2. 
After t=5, the area will be 100* 3.14 square= 314 square units.

3 0

3 years ago

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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 $\mu = 522, \sigma = 106$