25pt is it less than, greater than or equal to 13qt

Students at a bake sale sell bags of cookies for $2.35 each and bags of miniature muffins for $1.75 each. While selling their ba

inessss [21]

Answer:

Option C (The expression 2.35c + 1.75m represents the money earned from selling c bags of cookies and m bags of muffins).

Step-by-step explanation:

It is given that a bag of cookies is priced at $2.35 and a bag of muffin is priced at $1.75. It is also given that the student sells c bags of cookies and m bags of muffins. The total revenue earned by the student by selling c bag of cookies = $2.35 * c bags = $2.35c and the total revenue earned by the student by selling m bag of muffins = $1.75 * m bags = $1.75m. Therefore, the total revenue earned by the student by selling both cookies and muffins = 2.35c + 1.75m. This expression (2.35c + 1.75m) represents the money earned from selling c bags of cookies and m bags of muffins. Therefore, C is the correct choice!!!

3 0

3 years ago

A variety of stores offer loyalty programs. Participating shoppers swipe a bar-coded tag at the register when checking out and r

dsp73

Answer:

$t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

Now we can find the degrees of freedom:

$df=n-1=80-1=79$

Now we can calculate the p value with the alternative hypothesis using the following probability:

$p_v =P(t_{(79)}>2.236)=0.0141$

Using a significance level of 0.01 or 1% we have enough evidence to FAIL to reject the null hypothesis and we can conclude that shoppers participating in the loyalty program NOT spent more on average than typical shopper at this significance level assumed

Step-by-step explanation:

Information given

$\bar X=130$ represent the sample mean

$s=40$ represent the sample standard deviation

$n=80$ sample size

$\mu_o =120$ represent the value to verify

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if shoppers participating in the loyalty program spent more on average than typical shoppers (120) , the system of hypothesis would be:

Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

The statistic for this case is given by:

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

Replacing the info given we got:

$t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

Now we can find the degrees of freedom:

$df=n-1=80-1=79$

Now we can calculate the p value with the alternative hypothesis using the following probability:

$p_v =P(t_{(79)}>2.236)=0.0141$

Using a significance level of 0.01 or 1% we have enough evidence to FAIL to reject the null hypothesis and we can conclude that shoppers participating in the loyalty program NOT spent more on average than typical shopper at this significance level assumed

3 0

4 years ago

What is -2 1/2 + 1 1/3 in fraction form?

olga nikolaevna [1]

$-2\frac{1}{2} +1\frac{1}{3}=\frac{-1}{6}$