Carmen draws this area model to help her divide 1,800 by 9. She says the quotient is 20

Machines at a factory produce circular washers with a specified diameter. The quality control manager at the factory periodicall

swat32

Answer:

The Type I error that could result from the test is that The test provides convincing evidence that the proportion is greater than 90%, but the actual proportion is equal to 90%.

Step-by-step explanation:

We are given that machines at a factory produce circular washers with a specified diameter.

The quality control manager at the factory periodically tests a random sample of washers to be sure that greater than 90 percent of the washers are produced with the specified diameter.

Let p = proportion of all washers produced with the specified diameter

So, Null hypothesis, $H_0$ : p = 90%

Alternate Hypothesis, $H_A$ : p > 90%

Here, null hypothesis states that the proportion of all washers produced with the specified diameter is equal to 90%.

On the other hand, alternate hypothesis states that the proportion of all washers produced with the specified diameter is greater than 90%.

Now, Type I error states the Probability of rejecting null hypothesis given the fact that null hypothesis was true or in other words Probability of rejecting a true hypothesis.

So, according to our question Type I error would be that the test provides convincing evidence that the proportion is greater than 90%, but the actual proportion is equal to 90%.

0 0

2 years ago

23. A grid shows the positions of a subway stop and your house. The subway stop is located at (-5, 2) and your house is located

Sindrei [870]

(-5, 2)(-9,9)
D = √( (x2 - x1)^2 + (y2 - y1)^2 )
D = √( (-9+5)^2 + (9 - 2)^2 )
D =√( (-4)^2 + (7)^2 )
D = √( 16 + 49 )
D = √65
D = 8

7 0

2 years ago

Given the following formula, solve for v. s = 1/2a^2v + c

weeeeeb [17]

Since there is no initial condition, the exact value of v cannot be determined, but you can set the equation up for a general evaluation of the situation.

s = 1/2a^2*v+c

First you need to subtract c from both sides to get

s-c = 1/2a^2*v

then you can just divide both sides by 1/2a^2 to get v

(2(s-c))/a^2=v

when dividing a fraction, such as 1/2, make sure to keep in mind that you're really multiplying the flipped version, so that dividing by 1/2 means multiplying by 2.

7 0

2 years ago

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Translate “the sum of x and one-half of x” into mathematical expression

Aleks [24]

Answer:

X+ 0.5X or 1.5X

Step-by-step explanation:

the sum means that it want an addition problem. and so it is saying X + 1/2 (0.5)X which gives you 1.5X or you can leave it has X+0.5X

5 0

2 years ago

30 PTS!!!!! Cole needs to include at least 11 mg and at most 22 mg of iron in his diet each day. Cole plans to eat multiple ba

ser-zykov [4K]

Answer:

See explanation

Step-by-step explanation:

Part A: If Cole eats 3 bananas each day. These 3 bananas contain $3\cdot 0.3=0.9$ mg of iron.

Let x be the number of vitamins Cole takes. Each whole vitamin contains 4 mg of iron, then x vitamins contain 4x mg of iron.

The total amount of iron Cole takes is

$0.9+4x\ mg$

Cole needs to include at least 11 mg and at most 22 mg of iron in his diet each day, so

$11\le 0.9+4x\le 22$

Solve it:

$11-0.9\le 4x\le 22-0.9\\ \\10.1\le 4x\le 21.1\\ \\2.525\le x\le 5.275$

The smallest number of vitamins - 3, the greatest number of vitamins - 5.

Part B: Cole continues to eat 3 bananas each day. He pays $\$2\cdot 7=\$14$ for bananas per week.

If Cole spends a total of $20 on bananas and vitamins this week, then he spent $20 - $14 = $6 on vitamins.

Let x be the number of vitamins Cole takes this week.

First 10 vitamins are free, then x - 10 vitamins each cost $0.50 (he pays an additional dollar for every 2 additional vitamins he purchases) and they cost

$\$0.50(x-10)$

This amount of money is exactly $6, so

$0.50(x-10)=6\\ \\5(x-10)=60\\ \\x-10=12\\ \\x=22$

This means Cole takes 3 vitamins per day (the whole number) and he is able to meet his iron requirement every day this week, based on answer in Part A.

5 0

2 years ago

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