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n200080 [17]
2 years ago
13

Please help The factors of -18 are listed in the table …

Mathematics
1 answer:
pav-90 [236]2 years ago
8 0

Answer:

x= -2 and 9

Step-by-step explanation:

x ^{2}  + 7x - 18

factorize it

- 18 = 2×3×3

TO MAKE IT 7

3×3=9

9-2 =7

x ^{2}  + 9x - 2x - 18x = 0

x(x+9)-2(x+9)=0

x-2=0 ,x+9=0

x =2 x=-9

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Step-by-step explanation:

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If, triangle ABC, the measure of angle B is greater than 90 degrees, and AB=BC, what is a possible measure for angle C in degree
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A cognitive psychologist wants to know whether memory performance is changed by old age. She randomly selects 7 elderly individu
defon

Answer:

z = \frac{418.7-465.6}{\frac{53.6}{\sqrt{7}}}= -2.340

Now we can calculate the p value with the following probability

p_v =2*P(z

If we use the commonly significance level of 0.05 we see that the p value is lower than these values we can conclude that the true mean is different from the value of 465.6 at 5% of significance

Step-by-step explanation:

Data provided

\bar X=418.7 represent the mean score on the standardized memory test

\sigma=53.6 represent the population standard deviation

n=7 sample size    

\mu_o =465.6 represent the value that we want to verify

z would represent the statistic

p_v represent the p value

System of hypothesis

We are interested to check if the memory performance for elderly individuals differs from that of the general population (mean different from 465.6), the system of hypothesis are then:

Null hypothesis:\mu = 465.6    

Alternative hypothesis:\mu \neq 465.6    

Since we know the population deviation we can use the z statistic from the z test for the true mean:

z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}} (1)    

Replacing the info given we got:

z = \frac{418.7-465.6}{\frac{53.6}{\sqrt{7}}}= -2.340

Now we can calculate the p value with the following probability

p_v =2*P(z

If we use the commonly significance level of 0.05 we see that the p value is lower than these values we can conclude that the true mean is different from the value of 465.6 at 5% of significance

6 0
4 years ago
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