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ValentinkaMS [17]
3 years ago
9

Solve the given inequality and graph the solution on a number line -x/2 + 3/2 + < 5/2 .

Mathematics
2 answers:
liq [111]3 years ago
5 0
Answers in Image attached

Hope this helps

irina [24]3 years ago
5 0
Well, in that case, here goes...

The first thing I'd do is clear the fractions- that is, I'd multiply each of them by 2 because 2 is the lowest common denominator. 

That leaves you with -x + 3 < 5
Now solving isn't too bad...

-x + 3 - 3 < 5 - 3                 (subtract 3 from both sides)
-x < 2

The x isn't alone! That - is hanging on...so multiply both sides by -1 to get rid of it! 

-x * -1 < 2 * -1          WAIT! We multiplied by a negative!!
-x * -1 > 2 * -1          Had to flip that inequality ofver!

x > -2


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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.

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Which expression correctly simplifies (312415)56?
Andreyy89

Answer:

\sf C)\quad \large\text{$\dfrac{3^{\frac{5}{12}}}{4^{\frac{1}{6}}}$}

Step-by-step explanation:

Given expression:

\large\text{$\left(\dfrac{3^{\frac{1}{2}}}{4^{\frac{1}{5}}}\right)^\frac{5}{6}$}

\textsf{Apply exponent rule} \quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}:

\large\text{$\implies \dfrac{\left(3^{\frac{1}{2}}\right)^\frac{5}{6}}{\left(4^{\frac{1}{5}}\right)^\frac{5}{6}}$}

\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:

\large\text{$\implies \dfrac{3^{(\frac{1}{2}\times \frac{5}{6})}}{4^{(\frac{1}{5}\times \frac{5}{6})}}$}

Simplify:

\large\text{$\implies \dfrac{3^{\frac{5}{12}}}{4^{\frac{1}{6}}}$}

Learn more about exponent rules here:

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Answer:

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Step-by-step explanation:

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3 years ago
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