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3 years ago

12

PLEASE HELP 7TH GRADE MATH

2 answers:

kaheart [24]3 years ago

4 0

Answer:

Step-by-step explanation:

Number 5 is L because you have 18 coins but you only want to find the probability of the pennies. So take 3/18 and simplify to 1/6 by dividing both the numerator and the denominator by 3.

Number 6 can be (-6x - 3), just use the distributive property on the original expression.

Pie3 years ago

4 0

Answer:

5. I.

6. -6x - 3

Step-by-step explanation:

For number five you add up all the coins, which equals 18. Secondly, you see how many pennies there are. There are 3 so the answer is 3/18. Then you simpy into 1/6.

For number six, you want to multiply 3 by -2x and then multiply 3 by -1. You will get the answer -6x -3. If you see a number outside the parenthesis then multiply by the number inside of it.

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3 years ago

Heights of young adult women in the United States are approximately Normal with a mean of 64 inches and standard deviation 2.7 i

sweet-ann [11.9K]

Answer:

The proportion of all young adult women in the United States are taller than 6 feet(72 inches ) is

$P( X > 72) =0.0015233$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 64 \ inches$

The standard deviation is $\sigma = 2.7 \ inches$

Generally 6 feet is equivalent to 6 * 12 = 72 inches

Generally the proportion of all young adult women in the United States are taller than 6 feet(72 inches ) is mathematically represented as

$P( X > 72) = P(\frac{X - \mu }{\sigma } > \frac{72 - 64 }{ 2.7 } )$

$\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )$

$P( X > 72) = P(Z > 2.963 )$

From the z-table

The area under the normal curve to the right corresponding to 2.963 is

$P(Z > 2.963 ) = 0.0015233$

=> $P( X > 72) =0.0015233$

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3 years ago

Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypot

Nina [5.8K]

Answer: a. 0.1031; fail to reject the null hypothesis

Step-by-step explanation:

Given: Significance level : $\alpha=0.05$

The test statistic in a two-tailed test is z = -1.63.

The P-value for two-tailed test : $2P(Z>|z|)=2P(Z>|-1.63|)=0.1031$ [By p-value table]

Since, 0.1031 > 0.05

i.e. p-value > $\alpha$

So, we fail to reject the null hypothesis. [When p< $\alpha$ then we reject null hypothesis ]

So, the correct option is a. 0.1031; fail to reject the null hypothesis.

8 0

3 years ago