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likoan [24]
3 years ago
5

Find the value of x.

Mathematics
1 answer:
grandymaker [24]3 years ago
7 0
First do 70+24=94
Then do 180-94=86
Next 86+54=140
Then 180-140=40
Next 40+80=120
Lastly 180-120=60

X=60
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Ill mark you brainlist plz help if you put something random I will report you! I
Kisachek [45]

Answer:

x = 0

y = 2

Step-by-step explanation:

8 0
3 years ago
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<img src="https://tex.z-dn.net/?f=%5Coverline%7BAB%7D" id="TexFormula1" title="\overline{AB}" alt="\overline{AB}" align="absmidd
ad-work [718]

Answer:

B: 65°

Step-by-step explanation:

Step 1:

AB ║ ED      Given

Step 2:

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I need help its math
Degger [83]

Given: w²+9=6w [0, 3, 6, 9]

We need to substitute each number for w and find which works

Sub Zero: 0²+9=6×0

Simplify: 9=0. 9 does not equal 0, so it is not 0

Sub Three: 3²+9=6×3

Simplify: 9+9=18

Addition: 18=18

The statement 18=18 is true, therefore we have our answer.

Answer: The second option is correct. 3 would be the correct number to substitute to make the equation correct.

3 0
3 years ago
A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime
MrRissso [65]

Answer:

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Step-by-step explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let \mu = <u><em>true average percentage of organic matter</em></u>

So, Null Hypothesis, H_0 : \mu = 3%      {means that the true average percentage of organic matter in such soil is 3%}

Alternate Hypothesis, H_A : \mu \neq 3%      {means that the true average percentage of organic matter in such soil is something other than 3%}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

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

where, \bar X = sample mean percentage of organic matter = 2.481%

             s = sample standard deviation = 1.616%

            n = sample of soil specimens = 30

So, <u><em>the test statistics</em></u> =  \frac{2.481-3}{\frac{1.616}{\sqrt{30} } }  ~ t_2_9

                                     =  -1.76

The value of t-test statistics is -1.76.

(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have <u><em>sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have <u><em>insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

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A car travels at an average speed of 48 miles per hour. how long does it take to travel 204 miles?
Strike441 [17]
A car travels at an average speed of 48 miles per hour. how long does it take to travel 204 miles?It takes it 4.25 hours

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