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3 years ago

Your math Teacher has assigned homework in 9/10 of your last classes. Describe the likelihood of having homework tonight.

Mathematics

1 answer:

Solnce55 [7]3 years ago

4 0

Answer:

It is very likely that you will have homework tonight. Have fun! :)

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Cindy has a savings account with $7,500 in it that earns 7% simple interest per year.

WINSTONCH [101]

Answer:

The interest is $1050 and the amount is $8550.

Step-by-step explanation:

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3 years ago

How many ordered quadruples (w, x, y, z) of positive odd integers are there

guajiro [1.7K]

Answer:

165

Step-by-step explanation:

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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

4 0

3 years ago

Is 23,27,and 29 a prime number? Show ur work

Eduardwww [97]

23 is a prime number because it is only divisible by (has factors of) 1 and itself (23).
27 is NOT a prime number because it is divisible by more than 1 and itself (also 3 and 9).
29 is a prime number because it is only divisible by 1 and 29.

5 0

3 years ago

Read 2 more answers

day care center has 7 teacher for their after school program.currently 42 students are enrolled for a after school care at the s

DerKrebs [107]

Divide 7 into 42 your answer would be 6 then add 7 and get '13 $7 divide 42 equals 6 plus 7 equals '13$

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3 years ago