Converse means switching the hypothesis and conclusion of a conditional statement. The converse of x > y, is y > x. Therefore, the answer to your question is y > x, switching the hypothesis and conclusion of the statement.

5 0

3 years ago

Honon spent 3 1/4 hours on homework while Sequoia spent 2 5/6 hours on homework. How much more time did Honon spend on homework

Rzqust [24]

Answer:

5/12 hours

Step-by-step explanation:

Honon = 3 1/4 hours

Sequoia = 2 5/6 hours

How much more time did Honon spend on homework than Sequoia

= number of hours Honon spent - number of hours Sequoia spent

= 3 1/4 hours - 2 5/6 hours

= 13/4 - 17/6

= (39-34) /12

= 5/12 hours

6 0

3 years ago

Calcium is an essential nutrient for strong bones and for controlling blood pressure and heart beat. Because most of the body’s

irakobra [83]

Answer:

No, because the 95% confidence interval contains the hypothesized value of zero.

Step-by-step explanation:

Hello!

You have the information regarding two calcium supplements.

X₁: Calcium content of supplement 1

n₁= 12

X[bar]₁= 1000mg

S₁= 23 mg

X₂: Calcium content of supplement 2

n₂= 15

X[bar]₂= 1016mg

S₂= 24mg

It is known that X₁~N(μ₁; σ²₁), X₂~N(μ₂;δ²₂) and σ²₁=δ²₂=?

The claim is that both supplements have the same average calcium content:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

α: 0.05

The confidence level and significance level are to be complementary, so if 1 - α: 0.95 then α:0.05

since these are two independent samples from normal populations and the population variances are equal, you have to use a pooled variance t-test to construct the interval:

[(X[bar]₁-X[bar]₂) ± $t_{n_1+n_2-2;1-\alpha /2}$ * $Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$ ]

$t_{n_1+n_2-2;1-/2}= t_{25;0.975}= 2.060$

$Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} }= \sqrt{\frac{11*529+14*576}{12+15-2} } = 23.57$

[(1000-1016)±2.060*23.57* $\sqrt{\frac{1}{12} +\frac{1}{15} }$ ]

[-34.80;2.80] mg

The 95% CI contains the value under the null hypothesis: "zero", so the decision is to not reject the null hypothesis. Then using a 5% significance level you can conclude that there is no difference between the average calcium content of supplements 1 and 2.

I hope it helps!

3 0

3 years ago