Tii

Tariq lives in Palm Beach, Florida, and purchased a house for $250,000. The amount of millage for property tax is 11.8. What is

IRISSAK [1]

Answer:

$29500

Step-by-step explanation:

We have been given that Tariq lives in Palm Beach, Florida, and purchased a house for $250,000. The amount of millage for property tax is 11.8%. We are asked to find the amount of property tax that Tariq paid.

To find the amount of property tax paid by Tariq, we need to calculate 11.8% of $250,000.

$\text{Amount of property tax paid by Tariq}=\$250,000\times \frac{11.8}{100}$

$\text{Amount of property tax paid by Tariq}=\$2500\times 11.8$

$\text{Amount of property tax paid by Tariq}=\$29,500$

Therefore, Tariq paid an amount of $29500 in tax.

6 0

3 years ago

Rihanna bought 3 pounds of oranges for $2.52. How much did she pay per pound?

PIT_PIT [208]

which is 0.78

3 divided by 2.52

5 0

3 years ago

Special air bags are used to protect scientific equipment when a rover lands on the surface of Mars. On Earth, the function appr

PSYCHO15rus [73]

The nearest tenth of how fast a rover will hit Mars' surface after a bounce of 15 ft in height is 20.7ft/s.

<h3>What is the approximation about?</h3>

From the question:

Mars: F(x) = 2/3 $\sqrt{64x}$

Therefore, If x = 15

Then:

f (15) = 2/3 $\sqrt[8]{15}$

= 16/3 $\sqrt{15}$

= 20.7ft/s

Hence, The nearest tenth of how fast a rover will hit Mars' surface after a bounce of 15 ft in height is 20.7ft/s.

Learn more about Estimation from

brainly.com/question/24592593

#SPJ1

8 0

2 years ago

Mary open a bank account and deposits $16.00. Every month she deposits $30.00. Which expression shows the amount of money in her

NikAS [45]

Multiply the amount she deposits every month by the number of months (m) and add it to her starting amount:

The equation will look like this:

y = 30m + 16

4 0

4 years ago

Read 2 more answers

Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l

laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

Let $\mu$ = mean lead concentration for all such medicines.

So, Null Hypothesis, $H_0$ : $\mu$ = 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, $H_A$ : $\mu$ < 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$ ~ $t_n_-_1$

where, $\bar X$ = sample mean lead concentration = $\frac{\sum X}{n}$ = 12.25 mu

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 6.96 mu

n = sample size = 10

So, test statistics = $\frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}$ ~ $t_9$

= -2.16

The value of t test statistics is -2.16.

Now, the P-value of the test statistics is given by the following formula;

P-value = P( $t_9$ < -2.16) = 0.031

Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test. Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

7 0

4 years ago