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skelet666 [1.2K]
3 years ago
5

Steve,Jerry and Ron were paid $29.25 to remove garden gnomes. They each worked 4 hours, exempt for Ron, who was 1 hour late. How

much of the 29.25 should Ron receive?
Mathematics
1 answer:
Alenkasestr [34]3 years ago
5 0
Ron should receive $7.98. Jerry and Steve each worked for 4 hours, while Ron was one hour late, so he worked for 3 hours. Together, they worked for 4*2+3=11 hours, receiving $29.25 in total. Each hour should be paid 29.25/11=2.66 dollars. Ron worked for three hours, so he should receive 3*2.66=7.98 dollars. 
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Please help me! Determine the area
Ghella [55]

9514 1404 393

Answer:

  31.41 ft²

Step-by-step explanation:

Heron's formula is useful when you have the three side lengths.

  A = √(s(s -a)(s -b)(s -c)) . . . . sides are a, b, c and s = (a+b+c)/2

Using the given side lengths, we have ...

  s = (8 +8.4 +13.5)/2 = 29.9/2 = 14.95

  A = √(14.95(14.95 -8)(14.95 -8.4)(14.95 -13.5)) = √(14.95×6.95×6.55×1.45)

  A = √986.81399375 ≈ 31.41 . . . . square feet

7 0
2 years ago
Down the Drain: In a typical 5-minute shower, 30 gallons of water flow down the drain. How much water is used after 10 minutes?​
Bogdan [553]

Answer: 60 gallons

Step-by-step explanation:

5 minute = 30 gallons

10 minutes will use double as m uch water because it's double as much time

Therefore, it will use 60 gallons of water

6 0
2 years ago
Use a linear approximation (or differentials) to estimate the given number. (Round your answer to five decimal places.) 3 217
Soloha48 [4]

Answer:

f(216) \approx 6.0093

Step-by-step explanation:

Given

\sqrt[3]{217}

Required

Solve

Linear approximated as:

f(x + \triangle x) \approx f(x) +\triangle x \cdot f'(x)

Take:

x = 216; \triangle x= 1

So:

f(x) = \sqrt[3]{x}

Substitute 216 for x

f(x) = \sqrt[3]{216}

f(x) = 6

So, we have:

f(x + \triangle x) \approx f(x) +\triangle x \cdot f'(x)

f(215 + 1) \approx 6  +1 \cdot f'(x)

f(216) \approx 6  +1 \cdot f'(x)

To calculate f'(x);

We have:

f(x) = \sqrt[3]{x}

Rewrite as:

f(x) = x^\frac{1}{3}

Differentiate

f'(x) = \frac{1}{3}x^{\frac{1}{3} - 1}

Split

f'(x) = \frac{1}{3} \cdot \frac{x^\frac{1}{3}}{x}

f'(x) = \frac{x^\frac{1}{3}}{3x}

Substitute 216 for x

f'(216) = \frac{216^\frac{1}{3}}{3*216}

f'(216) = \frac{6}{648}

f'(216) = \frac{3}{324}

So:

f(216) \approx 6  +1 \cdot f'(x)

f(216) \approx 6  +1 \cdot \frac{3}{324}

f(216) \approx 6  + \frac{3}{324}

f(216) \approx 6  + 0.0093

f(216) \approx 6.0093

6 0
3 years ago
What is the solution to the following equation?
lions [1.4K]

Answer:

D. x = 7

Step-by-step explanation:

NOTE : <u><em>there should be an equal sign somewhere in the given expression.</em></u>

suppose the equation is the following:

3(x-4)-5 = x-3

………………………………………………………

3(x - 4) - 5 = x - 3

⇔ 3x - 12 - 5 = x - 3

⇔ 3x - 17 = x - 3

⇔ 3x - 17 + 17= x - 3 + 17

⇔ 3x = x + 14

⇔ 2x = 14

⇔ x = 14/2

⇔ x = 7

8 0
2 years ago
Extreme cold and hot temperatures are known to affect the operation of electronic components. Winter is approaching and you are
Musya8 [376]

Answer:

A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F

Test statistic(Z) = 2.31

P-value = 0.0104

Step-by-step explanation:

Step 1

Null hypothesis: The average damaging temperature of nine iPods is 5°F

Alternate hypothesis: The average damaging temperature of nine iPods differs from 5°F

Step 2

Mean=5°F, Sd=3, df=n-1=9-1=8

The t-value corresponding to 8 degrees of freedom and 95% confidence level (5% significance level) is 2.306

Confidence Interval(CI) = (mean + or - t×sd/√n)

CI = (5 + 2.306×3/√8) = 5 + 6.918/2.828= 5+2.45=7.45°F

CI = (5 - 2.306×3/√8) = 5-2.45 = 2.55°F

Z = (sample mean - population mean)/(sd÷√n) = (5-2.55)/(3÷√8) = 2.45/1.061 = 2.31

Step 3

Using the standard distribution table, the cumulative area to the left of Z = 2.31 is 0.9896

P-value = 1 - 0.9896 = 0.0104

Step 4

Conclusion: A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F

7 0
3 years ago
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