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a_sh-v [17]
2 years ago
14

Which two numbers does the √12 fall between? √4 √9 √16 √25 To answer, type in the box below. To answer √4, you would type square

root 4.
Mathematics
2 answers:
Tcecarenko [31]2 years ago
5 0

Answer:

9 and 16...............................................


emmasim [6.3K]2 years ago
5 0
ANSWER

\sqrt{9} \: and \: \sqrt{16}

EXPLANATION

First, simplify the square roots.

\sqrt{4} = \sqrt{ {2}^{2} } = 2

\sqrt{9} = \sqrt{ {3}^{2} } = 3

\sqrt{16} = \sqrt{ {4}^{2} } = 4

\sqrt{25} = \sqrt{ {5}^{2} } = 5

Let us simplify t
\sqrt{12}
Although 12 is not a perfect square, it contains, a perfect square.

\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2 \sqrt{3}

We can now observe that,

3 \: < \: 2 \sqrt{3} \: < \: 4

Therefore,

\sqrt{9} \: < \: \sqrt{12} \: < \: \sqrt{16}

Hence

\sqrt{12}
is between,

\sqrt{9}
and

\sqrt{16}
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mezya [45]

Answer:

a. \frac{70}{100}

b. 0.7

c. 0.7=70%=\frac{70}{100}

Step-by-step explanation:

70% of 100 received an A, that is 70/100 times 100 equal to 70

\frac{70}{100}*100=70

a. 70% = \frac{70}{100}

b. \frac{70}{100} = 0.7

Notice that 70/100 is equal to 7/10 which is 0.7

c. 0.7=70%=\frac{70}{100}  

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3 years ago
Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled with weed and soil treat
stiks02 [169]

Answer:

Part A

b. 14.6 ± 7.38

Part B

b. 3.43

Part C

a. P-value < 0.01

Part D

b. There is sufficient evidence to reject the null hypothesis

Step-by-step explanation:

Part A

The given data are;

The number of seedlings in the field = 20

The number of seedlings selected to receive herbicide A = 10

The number of seedlings selected to receive herbicide B = 10

The height in centimeters of seedlings treated with Herbicide A, \overline x _1 = 94.5 cm

The standard deviation, s₁ = 10 cm

The height in centimeters of seedlings treated with Herbicide B, \overline x _2 = 109.1 cm

The standard deviation, s₂ = 9 cm

The 90% confidence interval for μ₂ - μ₁, is given as follows;

\left (\bar{x}_{2}- \bar{x}_{1}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;

The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18

α = 100% - 90% = 10%

∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05

t_{(0.025, \, 18)} = 1.734

C.I. = \left (109.1- 94.5 \right )\pm 1.734 \times \sqrt{\dfrac{10^{2}}{10}+\dfrac{9^{2}}{10}}

C.I. ≈ 14.6 ± 7.37714603353

The 90% C.I. ≈ 14.6 ± 7.38

b. 14.6 ± 7.38

Part B

With the hypotheses are given as follows;

H₀; μ₂ - μ₁ = 0

Hₐ; μ₂ - μ₁ ≠ 0

The two sample t-statistic is given as follows;

t=\dfrac{(\bar{x}_{2}-\bar{x}_{1})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}+\dfrac{s _{2}^{2}}{n_{2}}}}

t-statistic=\dfrac{(109.1-94.5)}{\sqrt{\dfrac{10^{2} }{10}+\dfrac{9^{2}}{10}}} \approx 3.43173361147

The two sample t-statistic ≈ 3.43

b. 3.43

Part C

From the t-table, the p-value, we have, the p-value < 0.01

a. P-value < 0.01

Part D

Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis

b. There is sufficient evidence to reject the null hypothesis.

4 0
2 years ago
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Answer:

C 117

Step-by-step explanation:

<2 = 63

<4 = <2  vertical angles

<4 + <7 = 180

63 + <7 = 180

subtract 63 from each side

<7 = 117


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