732,178+167-642,137=please evaluate

An article retails for 27.50. The city sales tax is 6%, and the federal excise tax is 8%. How much does the article cost altoget

-Dominant- [34]

Cost of the article = $27.50
Amount of city sales tax = (27.50 x 6%) = $1.65
Amount of excise tax = (27.50 x 8%) = $2.20
From my knowledge, we pay city sales tax and federal excise tax when we buy something, so the total cost would be $31.35

5 0

3 years ago

Read 2 more answers

Someone help!!! asap!!!!

pentagon [3]

Answer:

Hope it helps plz give BRAINLIES! :D

Step-by-step explanation:

$80

4 0

3 years ago

If instead of randomly selecting two people there were five, what is the probability that all five are right-handed

lozanna [386]

Answer:

1/32

Step-by-step explanation:

We can break this question down. If we had 1 person and they are right-handed, the probability is 1/2 because they are either left-handed or right-handed. Knowing that, we can multiply the probability of having 1 person being right-handed 5 times because there are 5 people in the question.

1/2 × 1/2 × 1/2 × 1/2 × 1/2 = 1/(2⁵) = 1/32

Therefore, the probability of randomly selecting 5 people who are right-handed is 1/32.

I hope this helps and please mark me as brainliest!

5 0

2 years ago

X+5=121 please help me thank you. brainliest if helped

vampirchik [111]

$Solution, solve\:for\:x,\:x+5=121\quad :\quad x=116$

$Steps:$

$x+5=121$

$\mathrm{Subtract\:}5\mathrm{\:from\:both\:sides},\\x+5-5=121-5$

$\mathrm{Simplify},\\x=116$

$\mathrm{The\:Correct\:Answer\:is\:x=116}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

5 0

2 years ago

According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100:114 (46.7%

mojhsa [17]

Answer:

$z=\frac{0.42 -0.467}{\sqrt{\frac{0.467(1-0.467)}{150}}}=-1.154$

$p_v =2*P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of girls born is not significantly different from 0.467

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=63 represent the number of girls born

$\hat p=\frac{63}{150}=0.42$ estimated proportion of girls born

$p_o=0.467$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion if girls is 0.467.:

Null hypothesis: $p=0.467$

Alternative hypothesis: $p \neq 0.467$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.42 -0.467}{\sqrt{\frac{0.467(1-0.467)}{150}}}=-1.154$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of girls born is not significantly different from 0.467

3 0

3 years ago