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erik [133]
2 years ago
8

Part of the population of 7000 elk at a wildlife preserve is infected with a parasite. A random sample of 50 elk shows that 8 of

them are infected. How many elk are likely to be infected
Mathematics
1 answer:
Nikolay [14]2 years ago
5 0

Answer:

1120

Step-by-step explanation:

8/50=x/7000

50x=56000

Divide 50 on both sides

1120

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The diagram below shows scalene triangle WXY. The measure of WXY is 71 degree.
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A  AND D

Step-by-step explanation:

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Five times the sum of e and 4 is equal to -7.
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5+ 4e = -7 is the correct answer.
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Add. ˉ8 + ˉ6 = ____
Nutka1998 [239]
The answer is D +14 because to negatives make a positive
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The number of cups of sugar (y) in a cookie recipe is proportional to the number of cups of flour (x). The recipe calls for 1.25
8090 [49]

Answer:

Step-by-step explanation:

Y/X = 1.25/1 so place the amount of flour required in the denominator and multiply the denominator amount by 1.25 to find the amount of sugar

This problem appears to be missing equation choices.

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2 years ago
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One of your peers claims that boys do better in math classes than girls. Together you run two independent simple random samples
JulijaS [17]

Answer:

Step-by-step explanation:

Hello!

To test if boys are better in math classes than girls two random samples were taken:

Sample 1

X₁: score of a boy in calculus

n₁= 15

X[bar]₁= 82.3%

S₁= 5.6%

Sample 2

X₂: Score in the calculus of a girl

n₂= 12

X[bar]₂= 81.2%

S₂= 6.7%

To estimate per CI the difference between the mean percentage that boys obtained in calculus and the mean percentage that girls obtained in calculus, you need that both variables of interest come from normal populations.

To be able to use a pooled variance t-test you have to also assume that the population variances, although unknown, are equal.

Then you can calculate the interval as:

[(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} }]

Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} } = \sqrt{\frac{14*(5.6^2)+11*(6.7^2)}{15+12-2} }= 6.108= 6.11

t_{n_1+n_2-2;1-\alpha /2}= t_{15+12-2;1-0.05}= t_{25;0.95}= 1.708

[(82.3-81.2) ± 1.708* (6.11*\sqrt{\frac{1}{15}+\frac{1}{12}  }]

[-2.94; 5.14]

Using a 90% confidence level you'd expect the interval [-2.94; 5.14] to contain the true value of the difference between the average percentage obtained in calculus by boys and the average percentage obtained in calculus by girls.

I hope this helps!

3 0
2 years ago
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