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3 years ago

(Please help, Very desperate and I have a limited amount of time to get this done please hurry asap)

Mathematics

1 answer:

Romashka [77]3 years ago

5 0

Answer:

Pic1 answers:

P(blue) = 4/26,

P(yellow or green) = 14/26

P(purple) = 0.0%

P(red or green) = 18/26

P(not pink) = 26/26

P(not blue) = 22/26

Pic2 answers:

P(u) = 1/12

P(Vowel) = 6/12

P(a,b,c,d,e) = 4/12

P(Not A) = 10/12

P(vowel or consonant) = 12/12

P(not an m,s,t) = 9/12

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Hello,

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An 18 foot tall pole casts a shadow that is 28.8 feet long. At the same time a woman casts a shadow that is 8.8 feet long how ta

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3 years ago

What is the total probability of ending up with 3 blue marbles out of 7 marbles from the bag?

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3 years ago

. You are interested in the amount of time teenagers spend weekly working at part-time jobs. A random sample of 15 teenagers was

agasfer [191]

Answer:

$t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115$

Now we can calculate the degrees of freedom

$df=n-1=15-1=14$

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got $t_{critc}= 1.761$

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs

Step-by-step explanation:

Information given

$\bar X=147.3$ represent the sample mean for the amount of time spent at part time jobs

$s=50$ represent the sample standard deviation

$n=15$ sample size

$\mu_o =120$ represent the value to check

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to analyze if the true mean for the amount of time spent at part time jobs is higher than 120, the system of hypothesis would be:

Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

Since we don't know the population deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115$

Now we can calculate the degrees of freedom

$df=n-1=15-1=14$

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got $t_{critc}= 1.761$

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs

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3 years ago