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EleoNora [17]
3 years ago
13

Could someone help a girl out?

Mathematics
2 answers:
erik [133]3 years ago
8 0

Answer:

Yes I would have to agree with Goliath, x = 19.

Step-by-step explanation:

You can mark Goliath Brainlest if you want they did the answer first.

Rzqust [24]3 years ago
3 0

Answer:

x = 19

Step-by-step explanation:

Trapezium area formula: \frac{a+b}{2} * h, where a & b are the two bases.

Now, if we plug in our known values, we can solve for the missing base.

\frac{5 + x}{2}  * 4 = 48

\frac{5 + x}{2} = 12\\\\5 + x = 24\\\\x = 19

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Answer:

y = 16sin(2π/ 3)t.

Step-by-step explanation:

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The formula is y = A sin(2πft)  where A = the amplitude, f = frequency which  is 1/period and t = the time.

So for this SHM the function is

y = 16sin(2π/ 3) t.

You are correct.

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Independent random samples from two regions in the same area gave the following chemical measurements (ppm). Assume the populati
Basile [38]

Answer:

d. The interval contains only negative numbers. We cannot say at the required confidence level that one region is more interesting than the other.

Step-by-step explanation:

Hello!

You have the data of the chemical measurements in two independent regions. The chemical concentration in both regions has a Gaussian distribution.

Be X₁: Chemical measurement in region 1 (ppm)

Sample 1

n= 12

981 726 686 496 657 627 815 504 950 605 570 520

μ₁= 678

σ₁= 164

Sample mean X[bar]₁= 678.08

X₂: Chemical measurement in region 2 (ppm)

Sample 2

n₂= 16

1024 830 526 502 539 373 888 685 868 1093 1132 792 1081 722 1092 844

μ₂= 812

σ₂= 239

Sample mean X[bar]₂= 811.94

Using the information of both samples you have to determina a 90% CI for μ₁ - μ₂.

Since both populations are normal and the population variances are known, you can use a pooled standard normal to estimate the difference between the two population means.

[(X[bar]₁-X[bar]₂)±Z_{1-\alpha /2}* \sqrt{\frac{Sigma^2_1}{n1}+\frac{Sigma^2_2}{n_2}  }]

Z_{1-\alpha /2}= Z_{0.95}= 1.648

[(678.08-811.94)±1.648*\sqrt{\frac{164^2}{12}+\frac{239^2}{16}  }]

[-259.49;-8.23]ppm

Both bonds of the interval are negative, this means that with a 90% confidence level the difference between the population means of the chemical measurements of region 1 and region 2 may be included in the calculated interval.

You cannot be sure without doing a hypothesis test but it may seem that the chemical measurements in region 1 are lower than the chemical measurements in region 2.

I hope it helps!

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