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bazaltina [42]
3 years ago
6

Can u help me plssss

Mathematics
2 answers:
Marysya12 [62]3 years ago
4 0
The answer couldn’t possibly be 120
Anastaziya [24]3 years ago
3 0

Answer:

120 I think.

Step-by-step explanation:

Because 10*12=120

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The average human gestation period is 270 days with a standard deviation of 9 days. The period is normally distributed. What is
Anna35 [415]

Answer:

Probability that a randomly selected woman's gestation period will be between 261 and 279 days is 0.68.

Step-by-step explanation:

We are given that the average human gestation period is 270 days with a standard deviation of 9 days. The period is normally distributed.

Firstly, Let X = women's gestation period

The z score probability distribution for is given by;

         Z = \frac{ X - \mu}{\sigma} ~ N(0,1)

where, \mu = average gestation period = 270 days

            \sigma = standard deviation = 9 days

Probability that a randomly selected woman's gestation period will be between 261 and 279 days is given by = P(261 < X < 279) = P(X < 279) - P(X \leq 261)

         P(X < 279) = P( \frac{ X - \mu}{\sigma} < \frac{279-270}{9} ) = P(Z < 1) = 0.84134

         P(X \leq 261) = P( \frac{ X - \mu}{\sigma} \leq \frac{261-270}{9} ) = P(Z \leq -1) = 1 - P(Z < 1)

                                                           = 1 - 0.84134 = 0.15866

<em>Therefore, P(261 < X < 279) = 0.84134 - 0.15866 = 0.68</em>

Hence, probability that a randomly selected woman's gestation period will be between 261 and 279 days is 0.68.

3 0
3 years ago
The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.10
Salsk061 [2.6K]

For the bell-shaped graph of the normal distribution of weights of Hershey kisses, the area under the curve is 1, the value of the median and mode both is 4.5338 G and the value of variance is 0.0108.

In the given question,

The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.1039 G.

We have to find the answer of many question we solve the question one by one.

From the question;

Mean(μ) = 4.5338 G

Standard Deviation(σ) = 0.1039 G

(a) We have to find for the bell-shaped graph of the normal distribution of weights of Hershey kisses what is the area under the curve.

As we know that when the mean is 0 and a standard deviation is 1 then it is known as normal distribution.

So area under the bell shaped curve will be

\int\limits^{\infty}_{-\infty} {f(x)} \, dx= 1

This shows that that the total area of under the curve.

(b) We have to find the median.

In the normal distribution mean, median both are same. So the value of median equal to the value of mean.

As we know that the value of mean is 4.5338 G.

So the value of median is also 4.5338 G.

(c) We have to find the mode.

In the normal distribution mean, mode both are same. So the value of mode equal to the value of mean.

As we know that the value of mean is 4.5338 G.

So the value of mode is also 4.5338 G.

(d) we have to find the value of variance.

The value of variance is equal to the square of standard deviation.

So Variance = (0.1039)^2

Variance = 0.0108

Hence, the value of variance is 0.0108.

To learn more about normally distribution link is here

brainly.com/question/15103234

#SPJ1

3 0
1 year ago
Plz help geometry will mark brainliest
stellarik [79]
15+16+10 =41
BH= 41
Bd=15
Df=16
Fh=10

Plz mark me brainalist answer
3 0
2 years ago
PLEASE HELP! 28 POINTS!
ehidna [41]
1.) 7.2 2.) 15 3.)12
4.) 23 5.) 100, 100, 1000
6 0
2 years ago
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The measure of a circumscribed angle is equal to 180° minus the measure of the __________
ss7ja [257]
I believe it is an inscribed angle but I am not positive
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3 years ago
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