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masya89 [10]
2 years ago
8

Suppose that normal human body temperatures are normally distributed with a mean of 37°C and a standard deviation of 0.2°C. The

percent of humans have a temperature below 36.8°C is what?
Mathematics
2 answers:
saveliy_v [14]2 years ago
4 0

Answer:

Step-by-step explanation:

The answer is not 15.87%

astra-53 [7]2 years ago
4 0

Answer:

16

Step-by-step explanation:

I did it and got it wrong it tells you the answer after the 2nd click :)

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The starting salary for a particular job is 1.2 million per annum. The salary increases each year by 75000 to a maximum of 1.5mi
Vlada [557]
<h2>Answer:</h2>

In the 5th year

<h2>Step-by-step explanation:</h2>

For the first year, the salary is 1.2million = 1,200,000

For the second year, the salary is 1.2million + 75000 = 1,200,000 + 75,000 = 1,275,000

.

.

.

For the last year, the salary is 1.5million = 1,500,000

This gives the following sequence...

1,200,000 1,275,000  .   .   . 1,500,000

This follows an arithmetic progression with an increment of 75,000.

<em>Remember that,</em>

The last term, L, of an arithmetic progression is given by;

L = a + (n - 1)d           ---------------(i)

<em>Where;</em>

a = first term of the sequence

n = number of terms in the sequence (which is the number of years)

d = the common difference or increment of the sequence

<em>From the given sequence,</em>

a = 1,200,000                          [which is the first salary]

d = 75,000                               [which is the increment in salary]

L = 1,500,000                          [which is the maximum salary]

<em>Substitute these values into equation (i) as follows;</em>

1,500,000 = 1,200,00 + (n - 1) 75,000

1,500,000 - 1,200,000 = 75,000(n-1)

300,000 = 75,000(n - 1)

\frac{300,000}{75,000} = n - 1

4 = n - 1

n = 5

Therefore, in the 5th year the maximum salary will be reached.

3 0
2 years ago
Plz help and if you don't answer correctly i will report
KIM [24]

Answer:

1st, 3rd, 4th, 2nd

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Suppose X is the breaking strength (newtons) of a material, and X is normally distributed with
Veronika [31]

Answer:

a) 0.997 is the  probability that the breaking strength is at least 772 newtons.

b) 0.974  is the probability that this material has a breaking strength of at least 772 but not more  than 820    

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 800 newtons

Standard Deviation, σ = 10 newtons

We are given that the distribution of  breaking strength is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P( breaking strength of at least 772 newtons)

P(x \geq 772)

P( x \geq 772) = P( z \geq \displaystyle\frac{772 - 800}{10}) = P(z \geq -2.8)

= 1 - P(z

Calculation the value from standard normal z table, we have,  

P(x \geq 772) = 1 - 0.003 = 0.997 = 99.7\%

0.997 is the  probability that the breaking strength is at least 772 newtons.

b) P( breaking strength of at least 772 but not more  than 820)

P(772 \leq x \leq 820) = P(\displaystyle\frac{772 - 800}{10} \leq z \leq \displaystyle\frac{820-800}{10}) = P(-2.8 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2.8)\\= 0.977 - 0.003 = 0.974 = 97.4\%

P(772 \leq x \leq 820) = 97.4\\%

0.974  is the probability that this material has a breaking strength of at least 772 but not more  than 820.

7 0
2 years ago
The graph of the function f(x) = (x + 6)(x + 2) is shown. Which statements describe the graph? Check all that apply.
Ira Lisetskai [31]
The answers to the question would be number 2, number 3, and number 5
5 0
2 years ago
Read 2 more answers
The Fortune 500 is comprised of Fortune magazine's annual list of the finite population of the top 500 companies in the U.S. ran
Serjik [45]

Answer:

The standard error of the proportion is 0.0367.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the standard error is s = \sqrt{\frac{p(1-p)}{n}}

In this question:

p = 0.16, n = 100

So

s = \sqrt{\frac{0.16*0.84}{100}} = 0.0367

The standard error of the proportion is 0.0367.

4 0
2 years ago
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