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butalik [34]
3 years ago
12

Which set of ordered pairs in the form of (x,y) does not represent a function of x? {(-1,2), (3,-2),(0,1),(5,2}

Mathematics
1 answer:
serg [7]3 years ago
5 0
For a relation to be a function it must be one-to-one or many-to-one.

One-to-many relation is not a function.

In the above options, C is One-to-many relation therefore cannot be a function.

The reason is that 3 alone maps onto -2 and 5, in the ordered pairs (3,-2) and (3,5). This disqualifies it from being a function.

Hence the graph of this relation will not pass the vertical line test

The correct answer is C
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The answer is C. Hope this helps
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A cola-dispensing machine is set to dispense 8 ounces of cola per cup, with a standard deviation of 1.0 ounce. The manufacturer
pshichka [43]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: ounces per cup dispensed by the cola-dispensing machine.

The population mean is known to be μ= 8 ounces and its standard deviation σ= 1.0 ounce. Assuming the variable has a normal distribution.

A sample of 34 cups was taken:

a. You need to calculate the Z-values corresponding to the top 5% of the distribution and the lower 5% of it. This means you have to look for both Z-values that separates two tails of 5% each from the body of the distribution:

The lower value will be:

Z_{o.o5}= -1.648

You reverse the standardization using the formula Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)

-1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }

X[bar]= 7.72ounces

The lower control point will be 7.72 ounces.

The upper value will be:

Z_{0.95}= 1.648

1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }

X[bar]= 8.28ounces

The upper control point will be 8.82 ounces.

b. Now μ= 7.6, considering the control limits of a.

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-7.6)/(1/√34))- P(Z≤7.72-7.6)/(1/√34))

P(Z≤7.11)- P(Z≤0.70)= 1 - 0.758= 0.242

There is a 0.242 probability of the sample means being between the control limits, this means that they will be outside the limits with a probability of 1 - 0.242= 0.758, meaning that the probability of the change of population mean being detected is 0.758.

b. For this item μ= 8.7, the control limits do not change:

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-8.7)/(1/√34))- P(Z≤7.72-8.7)/(1/√34))

P(Z≤-2.45)- P(Z≤-5.71)=0.007 - 0= 0.007

There is a 0.007 probability of not detecting the mean change, which means that you can detect it with a probability of 0.993.

I hope it helps!

5 0
3 years ago
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The Damon family owns a large grape vineyard in western New York along Lake Erie. The grapevines must be sprayed at the beginnin
solniwko [45]

Answer:

Step-by-step explanation:

This is a test of 2 population proportions. Let 1 and 2 be the subscript for the the vines infested using Pernod 5 and vines infested using Action. The population proportion of the vines infested using Pernod 5 and vines infested using Action would be p1 and p2 respectively.

p1 - p2 = difference in the proportion of the vines infested using Pernod 5 and vines infested using Action.

The null hypothesis is

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p1 - p2 = 0

The alternative hypothesis is

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it is a two tailed test

Sample proportion = x/n

Where

x represents number of success(number of complaints)

n represents number of samples

For vines infested using Pernod 5,

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x2 = 39

n2 = 400

P2 = 39/400 = 0.098

The pooled proportion, pc is

pc = (x1 + x2)/(n1 + n2)

pc = (26 + 39)/(410 + 400) = 0.08

1 - pc = 1 - 0.08 = 0.92

z = (p1 - p2)/√pc(1 - pc)(1/n1 + 1/n2)

z = (0.063 - 0.098)/√(0.08)(0.92)(1/410 + 1/400) = - 0.035/0.019066

z = - 1.84

Since it is a two tailed test, the curve is symmetrical. We will look at the area in both tails. Since it is showing in one tail only, we would double the area

From the normal distribution table, the area below the test z score in the left tail 0.033

We would double this area to include the area in the right tail of z = 1.84 Thus

p = 0.033 × 2 = 0.066

By using the p value,

Since alpha, 0.01 < than the p value, 0.066, then we would fail to reject the null hypothesis.

Therefore, At a 1% level of significance, we cannot conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action

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Answer:

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Step-by-step explanation:

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