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2 years ago

Factor completely x2 + 8x - 20

2 answers:

8 0

Answer=(x-2) (x+10)

For this it helps if you do a factoring table but since I will not for the sake of not having an image to show you, I will try and explain my answer.

The expression is:

x²+8x-20

In this case, since the x² doesn't have a number before it, we know that the beginning x's are solo, so we have...

(x+?)(x+?)

Now to solve the rest of the expression, we need to find two numbers that when multiplied equal -20 and when added equal 8.

a*b=-20
a+b=8

Lets look at the factor pairs of -20

(-1,20) or (1,-20) or (-2,10) or (2,-10) or (-4,5) or (4,-5)

Of all of these, only one pair have a sum of 8

So we can solve the expression

x²+8x-20=(x-2)(x+10)

kvv77 [185]2 years ago

3 0

X-2
Hope this helps!
Vote me brainliest!

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Step-by-step explanation:

Hello!

A pilot study was conducted to test if three food rewards are equally appealing to the participants.

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16 preferred cupcakes (CC)

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The objective of this pilot study is to test if the observed frequencies follow a theoretical model/ distribution. To analyze this, you have to apply a Chi Square Goodness to Fit test. $X^2=sum \frac{(O_i-E_i)^2}{E_i} ~~X^2_{k-1}$ Where k= number of categories of the variable.

For this example the statistical hypotheses are:

H₀: P(CC)=P(CB)=P(DA)= 1/3

H₁: At least one of the proportions isn't equal to the others.

To calculate the expected frequencies for each category you have to use the formula: $E_i= n* P_i$ where Pi represents the theoretical proportion for the i category, stated in the null hypothesis.

$E_{CC}= n* P(CC)= 60*1/3= 20$

$E_{CB}= n*P(CB)= 60*1/3= 20$

$E_{DA}= n* P(DA)= 60* 1/3= 20$

The cutoff or critical value indicates the beginning of the rejection region for the hypothesis test. For the Chi-Square tests, the rejection region is always one-tailed to the right, meaning that you'll reject the null hypothesis if the value of the statistic is big. For the goodness to fit test you have k-1 degrees of freedom, so the critical value will be:

Assuming α: 0.05

$X^2_{k-1;1-\alpha /2}= X^2_{2;0.975}= 7.378$

The rejection region is then X²₂ ≥ 7.378

$X^2_{H_0}= \frac{(O_{CC}-E_{CC})^2}{E_{CC}} + \frac{(O_{CB}-E_{CB})^2}{E_{CB}} + \frac{(O_{DA}-E_{DA})^2}{E_{DA}} = \frac{(16-20)^2}{20} +\frac{(26-20)^2}{20} +\frac{(18-20)^2}{20}= \frac{14}{5}= 2.8$

I hope this helps!

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2 years ago