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3 years ago

I suck at math please help me

Mathematics

1 answer:

attashe74 [19]3 years ago

3 0

-2/15 is the answer

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Write an equation that you can use to solve for x. Enter your answer in the box.

N76 [4]

The answer is X=27 degrees

Steps-
90-63=27

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3 years ago

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leigh has a piece of rope that is 6 2/3 feet long. how do you write 6 2/3 as a fraction greater than 1?

AnnZ [28]

Answer:

$\frac{20}{3}\ ft$

Step-by-step explanation:

we have

$6\frac{2}{3}\ ft$

Convert mixed number to an improper fraction

Remember that

$6\frac{2}{3}=6+\frac{2}{3}=\frac{6*3+2}{3}=\frac{20}{3}\ ft$

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3 years ago

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Find the slope from the given points (-5,3) (2,4)

Alecsey [184]

Answer:1/7

(-5,3)= (x1,y1)

(2,4)= (x2,y2)

Slope= $\frac{y2-y1}{x2-x1}$

= $\frac{4-3}{2-(-5)}$

= $\frac{1}{7}$

4 0

3 years ago

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Martin charges $15 for every 5 bags of leaves he rakes. Last weekend, he raked 27 bags of leaves. How much money did he earn?

Burka [1]

Set up a proportion and then cross multiply to solve. OR. 15/5 = $3 per bag is the unit rate per 1 bag.
Multiply 27*3=$81

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4 years ago

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The attention span of children (ages 3 to 5) is claimed to be Normally distributed with a mean of 15 minutes and a standard devi

Agata [3.3K]

Answer:

If the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 minutes

Sample size, n = 10

Alpha, α = 0.05

Population standard deviation, σ = 4 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 15\text{ minutes}\\H_A: \mu > 15\text{ minutes}$

Since the population standard deviation is given, we use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Now, $z_{critical} \text{ at 0.05 level of significance for one tail} = 1.64$

Thus, we would reject the null hypothesis if the z-statistic is greater than this critical value.

Thus, we can write:

$z_{stat} = \displaystyle\frac{\bar{x} - 15}{\frac{4}{\sqrt{10}} } > 1.64\\\\\bar{x} - 15>1.64\times \frac{4}{\sqrt{10}}\\\bar{x} - 15>2.07\\\bar{x} > 15+2.07\\\bar{x} > 17.07$

Thus, the decision rule would be if the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

6 0

4 years ago