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Serggg [28]
3 years ago
11

2/-7 in standard form​

Mathematics
1 answer:
Elan Coil [88]3 years ago
5 0

Answer:

the answer is

x2 - 7 =0

Step-by-step explanation:

no explination

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find the length of the hypotenuse of a right triangle with legs of 12 ft and 16 ft. A 11 ft. B 18 ft. C 20 ft. D 28 ft
9966 [12]
C 20ft, shoukd be the answer.
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.

8 0
4 years ago
Find the prime factorizations foor the following numbers 36 and 84
Varvara68 [4.7K]
Answer:

36 = 2² * 3²
84 = 2² * 3 * 7


Explanation:

Pf of 36 = 2 * 2 * 3 * 3 which then becomes 2 (to the power of 2) times 3 (to the power of 2)

Pf of 84 = 2 * 2 * 3 * 7 which then becomes 2 (to the power of 2) * 3 * 7

If this helps it would be greatly appreciated if you thanked me.
6 0
4 years ago
Round 3,872.3155 to the nearest hundred
zlopas [31]

Answer:

3,872.32

Step-by-step explanation:

the nearest hundredth is 2 numbers on the right of the decimal sign. So you round up from 5 to get answer above.

6 0
4 years ago
Given angle A B C , in which Line AD bisects angle BAC. If m angle B=64 and m angle C =42, determine m angle BAD.
Andrew [12]
D=54 i might be wrong
3 0
4 years ago
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