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Shtirlitz [24]
3 years ago
13

What is the measure of ∠MIJ?

Mathematics
2 answers:
Tems11 [23]3 years ago
5 0
<MIJ = 1/2(360 - 2(55))
<MIJ = 1/2(360 - 110)
<MIJ = 1/2(250)
<MIJ = 125
answer is C. 125

mr Goodwill [35]3 years ago
5 0
I believe you would need to subtract 55 from 180 (the measure of a straight line)
180 - 55 = 125
125 is your answer

You could also solve by process of elimination
A and B can't be the answer as they are less than a 90-degree angle, and MIJ is clearly an obtuse angle (an angle greater than 90-degrees)
D can't be the answer as the 305 is greater than a straight angle (180)
So C is the only logical answer
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The average American man consumes 9.8 grams of sodium each day. Suppose that the sodium consumption of American men is normally
Alex Ar [27]

Answer:

(a) The distribution of <em>X</em> is <em>N</em> (9.8, 0.8²).

(b) The probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.

(c) The middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.

Step-by-step explanation:

The random variable <em>X</em> is defined as the amount of sodium consumed.

The random variable <em>X</em> has an average value of, <em>μ</em> = 9.8 grams.

The standard deviation of <em>X</em> is, <em>σ</em> = 0.8 grams.

(a)

It is provided that the sodium consumption of American men is normally distributed.

The random variable <em>X</em> follows a normal distribution with parameters <em>μ</em> = 9.8 grams and <em>σ</em> = 0.8 grams.

Thus, the distribution of <em>X</em> is <em>N</em> (9.8, 0.8²).

(b)

If X ~ N (µ, σ²), then Z=\frac{X-\mu}{\sigma}, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z ~ N (0, 1).

To compute the probability of  Normal distribution it is better to first convert the raw score (<em>X</em>) to <em>z</em>-scores.

Compute the probability that an American consumes between 8.8 and 9.9 grams of sodium per day as follows:

P(8.8

                           =P(-1.25

Thus, the probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.

(c)

The probability representing the middle 30% of American men consuming sodium between two weights is:

P(x_{1}

Compute the value of <em>z</em> as follows:

P(-z

The value of <em>z</em> for P (Z < z) = 0.65 is 0.39.

Compute the value of <em>x</em>₁ and <em>x</em>₂ as follows:

-z=\frac{x_{1}-\mu}{\sigma}\\-0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8-(0.39\times 0.8)\\x_{1}=9.488\\x_{1}\approx9.5     z=\frac{x_{2}-\mu}{\sigma}\\0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8+(0.39\times 0.8)\\x_{1}=10.112\\x_{1}\approx10.1

Thus, the middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.

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