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ddd [48]
3 years ago
11

Can someone help me with the question in the image?

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
3 0

Step-by-step explanation:

m< I = m < 2 = m < 3

7y+16 = 2X

7y - 2x +16 = O ............I

7y +16 = 4X _30

7y -4 x + 46 = 0 ...............2

solve eq I,2

2 X - 30= o

X= 15

substitute in I

7X= 14

y = 2

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Question<br> Simplify for a = 3, b = -4 and c = -1.<br><br> B^2(a - 2c)
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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
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