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prohojiy [21]
2 years ago
10

HI PLS HELP ME IF ITS LINEAR OR NONLINEAR

Mathematics
1 answer:
Sunny_sXe [5.5K]2 years ago
3 0

Answer:

nonlinear

Step-by-step explanation:

I checked but I think it's right

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Find the area and perimeter of the shape above
ELEN [110]

Answer:

Perimeter: 18.5

Area: 15

Step-by-step explanation:

P = a + b + c

P = 5 + 6 + 7.5

P= 18.5

A = 1/2bh

A = 1/2*7.5*4

A= 15

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What is 5pi/4 in degrees
Scilla [17]

5pi/4*180/pi= 225 degree

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3 years ago
Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. What perce
LenKa [72]

Answer:

14.28% of individual adult females have weights between 75 kg and 83 ​kg.

92.82% of the sample means are between 75 kg and 83 ​kg.

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}}.

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.

In this problem, we have that:

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. This means that \mu = 79, \sigma = 22.

What percentage of individual adult females have weights between 75 kg and 83 ​kg?

This percentage is the pvalue of Z when X = 83 subtracted by the pvalue of Z when X = 75. So:

X = 83

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

Z = \frac{83 - 79}{22}

Z = 0.18

Z = 0.18 has a pvalue of 0.5714.

X = 75

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

Z = \frac{75- 79}{22}

Z = -0.18

Z = -0.18 has a pvalue of 0.4286.

This means that 0.5714-0.4286 = 0.1428 = 14.28% of individual adult females have weights between 75 kg and 83 ​kg.

If samples of 100 adult females are randomly selected and the mean weight is computed for each​ sample, what percentage of the sample means are between 75 kg and 83 ​kg?

Now we use the Central Limit THeorem, when n = 100. So s = \frac{22}{\sqrt{100}} = 2.2.

X = 83

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

Z = \frac{83 - 79}{2.2}

Z = 1.8

Z = 1.8 has a pvalue of 0.9641.

X = 75

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

Z = \frac{75-79}{2.2}

Z = -1.8

Z = -1.8 has a pvalue of 0.0359.

This means that 0.9641-0.0359 = 0.9282 = 92.82% of the sample means are between 75 kg and 83 ​kg.

8 0
3 years ago
How do you divide 2 digit numbers by 2 digits number?
Novosadov [1.4K]
If the numbers were 50 divided by 12 this is how you would answer it. And I showed the reduced answer too.

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3 years ago
Simplify 12 to the 3rd over 12 to the 7th.
svetlana [45]

Answer

1,728 and 35,831,808

Step-by-step explanation:

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