What is 3.248 repeating as a simplified fraction?

If p=5 and q=6, evaluate the following expression:2p+3q/4

suter [353]

Answer:

7

Step-by-step explanation:

Substitute in numbers : 2(5)+3(6) /4

Combine Like Terms : 10+18/4 = 28/4

Answer : 7

8 0

3 years ago

A function graph that matches y=3x+1

Luda [366]

Answer:

Step-by-step explanation:

How is this question supposed to be answered with no specifications?

3 0

3 years ago

Justin had $24 to buy a gift card for his cousin Julie and he found a gift for $22 with 5% sales tax is added on will Justin hav

Aloiza [94]

Yes he would have enough money.

8 0

3 years ago

a poker hand of 5 cards is dealt from a standard 52 card deck. a full house consists of exactly 3 of one kind of card and a pair

sergiy2304 [10]

The triple card can be any one of 13 values (ace through king). For each of those, the double card can be any one of the remaining 12 values. So concerning only the face values of the cards, there are (13 x 12) = 156 different full hice. If you include attention to the suits, then there a lot more.

7 0

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