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3 years ago

Joe has eaten 3/5 of pizza. Jane has eaten 1/7 of pizza. how many times more pizza has Joe eaten than Jane

Mathematics

2 answers:

Lana71 [14]3 years ago

4 0

The answer would be 14/5.

First of all, get the denominators to be the same. The LCM is 5 × 7 = 3/5

So 1 /7 × 5/ 5 = 5 /3/5

And 2/ 5 × 7 /7 = 14 /35

Now we have 5 /35 and 14 /35

Then all you have to do is divide Joe's by Jane's to get the answer.

14 /35 ÷ 5 35 =14 /35 × 35/ 5 = 14 /5

Meaning the answer is 14/5.

Hope I could help! :)

Fynjy0 [20]3 years ago

4 0

well you need to subtract 3/5 and 1/7. how you do that is by finding the least common denominator (find the lowest number that can go into both numbers) and then subtract the nominators.

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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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