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3 years ago

10

complete the paragraph proof given: <1 and <2 are supplementary and <2 and <3 are supplementary Prove: <1 is cong

ruent to <3

1 answer:

Mice21 [21]3 years ago

3 0

<1 and <2 are supplementary; <2 and <3 are supplementary this is a given.

<1 and <3 are supplementary; Substitution Property

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John has two jobs. For daytime work at a jewelry store he is paid

djyliett [7]

Given Information:

John's mean monthly commission = μ = $10,000

Standard deviation of monthly commission = σ = $2,000

Answer:

$P(9,000 < X < 11,000) = 0.383\\\\P(9,000 < X < 11,000) = 38.3 \%$

The probability that John's commission from the jewelry store is between $9,000 and $11,000 is 38.3%

Step-by-step explanation:

What is Normal Distribution?

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.

We want to find out the probability that John's commission from the jewelry store is between $9,000 and $11,000?

$P(9,000 < X < 11,000) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(9,000 < X < 11,000) = P( \frac{9,000 - 10,000}{2,000} < Z < \frac{11,000 - 10,000}{2,000} )\\\\P(9,000 < X < 11,000) = P( \frac{-1,000}{2,000} < Z < \frac{1,000}{2,000} )\\\\P(9,000 < X < 11,000) = P( -0.5 < Z < 0.5 )\\\\P(9,000 < X < 11,000) = P( Z < 0.5 ) - P( Z < -0.5 ) \\\\$

The z-score corresponding to 0.50 is 0.6915

The z-score corresponding to -0.50 is 0.3085

$P(9,000 < X < 11,000) = 0.6915 - 0.3085 \\\\P(9,000 < X < 11,000) = 0.383\\\\P(9,000 < X < 11,000) = 38.3 \%$

Therefore, the probability that John's commission from the jewelry store is between $9,000 and $11,000 is 38.3%

How to use z-table?

Step 1:

In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 1.4, 2.2, 0.5 etc.)

Step 2:

Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 0.50 then go for 0.00 column)

Step 3:

Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.

4 0

3 years ago

Which of the following does NOT have a value of -5?

Contact [7]

Answer:

J. -13-18

Step-by-step explanation:

J will give the value -31

4 0

3 years ago

Amy has a piece of wood that measures 42 inches. The model shows the length remaining after she cut a piece from the 42-inch pie

Hitman42 [59]

Answer:

She has 42 pieces of wood each of 1 inch of length.

Step-by-step explanation:

Amy has 42 inches piece of wood.

She has to cut an inch.

After cutting pieces of inch each she counts the pieces to be 42.

Mathematically

Total length / unit lenght = Number of pieces

42 inches/ 1 inch= 42 pieces.

She has 42 pieces of wood each of 1 inch of length.

6 0

3 years ago

Given: NQ is an altitude of △MNP

Virty [35]

Answer:

Statement: Reason:

ΔNQM and ΔNQP are right triangle Definition of a right triangle

SinM = h/p and SinP = h/m Definition of sine ratio

p sin M = m sin P Substitution property of equality

p sin M / pm = msinP / pm Division property of equality.

8 0

3 years ago

Question 7 of 10

andriy [413]

Answer:

C. n = 90; p = 0.8

Step-by-step explanation:

According to the Central Limit Theorem, the distribution of the sample means will be approximately normally distributed when the sample size, 'n', is equal to or larger than 30, and the shape of sample distribution of sample proportions with a population proportion, 'p' is normal IF n·p ≥ 10 and n·(1 - p) ≥ 10

Analyzing the given options, we have;

A. n = 45, p = 0.8

∴ n·p = 45 × 0.8 = 36 > 10

n·(1 - p) = 45 × (1 - 0.8) = 9 < 10

Given that for n = 45, p = 0.8, n·(1 - p) = 9 < 10, a normal distribution can not be used to approximate the sampling distribution

B. n = 90, p = 0.9

∴ n·p = 90 × 0.9 = 81 > 10

n·(1 - p) = 90 × (1 - 0.9) = 9 < 10

Given that for n = 90, p = 0.9, n·(1 - p) = 9 < 10, a normal distribution can not be used to approximate the sampling distribution

C. n = 90, p = 0.8

∴ n·p = 90 × 0.8 = 72 > 10

n·(1 - p) = 90 × (1 - 0.8) = 18 > 10

Given that for n = 90, p = 0.9, n·(1 - p) = 18 > 10, a normal distribution can be used to approximate the sampling distribution

D. n = 45, p = 0.9

∴ n·p = 45 × 0.9 = 40.5 > 10

n·(1 - p) = 45 × (1 - 0.9) = 4.5 < 10

Given that for n = 45, p = 0.9, n·(1 - p) = 4.5 < 10, a normal distribution can not be used to approximate the sampling distribution

A sampling distribution Normal Curve

45 × (1 - 0.8) = 9

90 × (1 - 0.9) = 9

90 × (1 - 0.8) = 18

45 × (1 - 0.9) = 4.5

Now we will investigate the shape of the sampling distribution of sample means. When we were discussing the sampling distribution of sample proportions, we said that this distribution is approximately normal if np ≥ 10 and n(1 – p) ≥ 10. In other words

Therefore;

A normal curve can be used to approximate the sampling distribution of only option C. n = 90; p = 0.8

3 0

3 years ago