The point-slope equation of a line is....? HELP ME

MARK AS BRAINLEST! 5 + x = 5

n200080 [17]

Answer:

x = 0

Step-by-step explanation:

You know the answer to this because you know the identity element for addition is 0: 5 + 0 = 5.

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Or, you can make use of the addition property of equality and add -5 to both sides of the equation:

5 - 5 + x = 5 - 5

x = 0 . . . . . . . . . . simplify

3 0

3 years ago

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Ciara is making a new dance outfit for a dance performance in her high school. She needs 2 1/2 yards of fabric for the shawl. Sh

sergey [27]

Answer:

1 1/4 yards of fabric

Step-by-step explanation:

2 1/2+ 1 3/4=4 1/4

4 1/4 - 3 = 1 1/4

8 0

3 years ago

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Can someone check this for me thanks so much

Zinaida [17]

Answer:

that looks like it is right

3 0

3 years ago

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Square root symbol 640 to the nearest hundreth

Vitek1552 [10]

The square root of 640 is 25.29

5 0

3 years ago

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Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l

laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

Let $\mu$ = mean lead concentration for all such medicines.

So, Null Hypothesis, $H_0$ : $\mu$ = 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, $H_A$ : $\mu$ < 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$ ~ $t_n_-_1$

where, $\bar X$ = sample mean lead concentration = $\frac{\sum X}{n}$ = 12.25 mu

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 6.96 mu

n = sample size = 10

So, test statistics = $\frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}$ ~ $t_9$

= -2.16

The value of t test statistics is -2.16.

Now, the P-value of the test statistics is given by the following formula;

P-value = P( $t_9$ < -2.16) = 0.031

Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test. Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

7 0

3 years ago