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suter [353]
3 years ago
9

Students at Westbrook middle school hold green team meetings to plan ways to help protect the environment. This week there are 1

44 members at the meeting which is 90% of the total group?
Mathematics
1 answer:
ICE Princess25 [194]3 years ago
8 0

Answer:

160 members.

Step-by-step explanation:

Please consider the complete question.

Students at Westbrook Middle School hold green team meetings to plan ways to help  protect the environment. This week there are 144 members at the meeting, which is 90%  of the total group.  What is the total number of green team members?

Let x represent total number of green team members.

90% of total number of green team members would be \frac{90}{100}x=0.90x.

Now we will equate 90% of total number of green team members with 144 and solve for x as:

0.90x=144

\frac{0.90x}{0.90}=\frac{144}{0.90}

x=160

Therefore, the total number of green team members is 160.

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Answer:

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Step-by-step explanation:

-2(-2)-4

=4-4

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3 years ago
Which of the following distinctions does Russia enjoy?
kap26 [50]

Answer:

A distinctions does Russia enjoy is explained below in details.

Step-by-step explanation:

  • About 11 percent of Russia is tundra, marshy plain, a treeless The tundra is Russia's northernmost zone.
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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!

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Dahasolnce [82]
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