All categories

3 years ago

Factor. 18b−32 Enter your answer in the boxes. [ ]([ ])

2 answers:

3 0

Here, GCF of 18 & 32 is 2, so take it as common factor,
18b - 32 = 2(9b - 16)

In short, Your Answer would be: 2(9b - 16)

Hope this helps!

Oliga [24]3 years ago

3 0

18b-32
/2 /2

-2(-9b+16)

You might be interested in

For a brainlist pls answer <br> + accurate explanation

densk [106]

The answer is (-2, 1) Step-by-step explanation

4 0

3 years ago

PlES heLP mEh >:D FS:D>FSD:F> SD:F

VashaNatasha [74]

70 I believe is the correct answer

3 0

3 years ago

Read 2 more answers

Write the equation of a line with a slope of 3 and a y-intercept of 4. Do 5p

lana66690 [7]

Answer:

f(x) = 4 + 3x

Step-by-step explanation:

5 0

3 years ago

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 $\chi^2_{crit}=7.815$