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2 years ago

12

Solve for kkk. \dfrac{3}{k} = \dfrac{4}{5} k 3 = 5 4 start fraction, 3, divided by, k, end fraction, equals, start fraction,

4, divided by, 5, end fraction k =k=

2 answers:

sweet-ann [11.9K]2 years ago

5 0

Answer: $\dfrac{15}{4}$

Step-by-step explanation:

Given

$\Rightarrow \dfrac{3}{k}=\dfrac{4}{5}$

Apply cross-multiplication

$\Rightarrow K=3\times \dfrac{5}{4}=\dfrac{15}{4}\\\\\Rightarrow k=3\ \dfrac{3}{4}$

The value of k is $\dfrac{15}{4}\ \text{or}\ 3\ \dfrac{3}{4}$

harina [27]2 years ago

3 0

Answer:

15/4

Step-by-step explanation:

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Consider the following hypothesis test. H0: μ ≥ 55 Ha: μ < 55 A sample of 36 is used. Identify the p-value and state your con

Ket [755]

Answer:

Step-by-step explanation:

Given that:

$H_o: \mu \ge 55 \ \\ \\ H_1 : \mu < 55$

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$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{54-55}{\dfrac{5.3}{\sqrt{36}} }$

$Z = \dfrac{-1}{\dfrac{5.3}{6} }$

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= 35

Using the Excel Formula:

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b

For x = 53 and s = 4.6

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{53-55}{\dfrac{4.6}{\sqrt{36}} }$

$Z = \dfrac{-2}{\dfrac{4.6}{6} }$

Z = -2.6087

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Using the Excel Formula:

P-Value = T.DIST(-2.6087,35,1) =0.0066

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$Conclusion: \ There \ is \ sufficient \ evidence \ to \ conclude \ that$ $\mu < 55$

c)

For x = 56 and s = 5.0

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }$

$Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }$

Z = 1.2

degree of freedom = n - 1

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Using the Excel Formula:

P-Value = T.DIST(1.2,35,1) = 0.88009

Decision: p-value is greater than significance level; do not reject $\mathbf{H_o}$

$Conclusion: \ There \ is \ insufficient \ evidence \ to \ conclude \ that \$ $\mu < 55$

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