What is 7/9 divided by 1/3?<br>0 2<br>0 2 1/9<br>0 2 1/3<br>0 3

Michael consumes a 16-ounce bag of potato chips and 12-ounce bag of pretzels every day after school. How many more ounces of pot

Paladinen [302]

Every day, Michael consumes 4 more ounces than the pretzels. Multiply 5 by 4 to recieve the answer:

4 x 5 = 20

So the answer is A.

7 0

2 years ago

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6. A government's economic role is complicated by the fact that

user100 [1]

Answer:B

Step-by-step explanation:

5 0

3 years ago

What is the median of the following data values 98,102,53,80,22

timurjin [86]

Answer:

80

Step-by-step explanation:

8 0

3 years ago

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The dimensions of a square are altered so that one dimension is increased by 7 feet, while the other is decreased by 2 feet. The

dangina [55]

Answer:

64 ft^2

Step-by-step explanation:

(x+7)(x-2)=90

x^2-2x+7x-14=90

x^2+5x-104=0

factor

(x+13)(x-8)=0

x=-13

x=8 ft original square side

area=8x8=64

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3 years ago

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The Fish and Game Department stocked a lake with fish in the following proportions: 30% catfish, 15% bass, 40% bluegill, and 15%

vagabundo [1.1K]

Answer:

1) $\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313$

2) $p_v =P(\chi^2_{3}>16.313)=0.000978$

And we got the same decision reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts

The Chi-Square test of independence is used "to determine if there is a significant relationship between two nominal (categorical) variables". And is defined with the following statistic:

$\chi^2 =\sum_{i=1}^n \frac{(O-E)^2}{E}$

Where O rpresent the observed values and E the expected values.

State the null and alternative hypothesis

Null hypothesis: The distribution is 30% catfish, 15% bass, 40% bluegill, and 15% pike

Alternative hypothesis: The distribution is NOT 30% catfish, 15% bass, 40% bluegill, and 15% pike

The observed values are given by the table given:

Catfish =112, BAss = 95, Bluegill=210, Pike=83

Calculate the expected values

In order to calculate the expected values we can use the following formula for each cell of the table

$E = \% Grand total$

$E_{Catfish}=500*0.3=150$

$E_{Bass}=500*0.15=75$

$E_{Bluegill}=500*0.4=200$

$E_{Pike}=500*0.15=75$

Part 1: Calculate the statistic

$\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313$

$\chi^2 =16.313$

Calculate the critical value

First we need to calculate the degrees of freedom given by:

$df= (categories-1)=(4-1)= 3$

Since the confidence provided is 95% the significance would be $\alpha=1-0.95=0.05$ and we can find the critical value with the following excel code: "=CHISQ.INV(0.95,3)", and our critical value would be $\chi^2_{crit}=7.815$

We can calculate also the p value:

$p_v =P(\chi^2_{3}>16.313)=0.000978$

And we got the same decision reject the null hypothesis at 5% of significance.

5 0

3 years ago

What is 7/9 divided by 1/3?0 20 2 1/90 2 1/30 3​

What is 7/9 divided by 1/3?0 20 2 1/90 2 1/30 3