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Darya [45]
3 years ago
9

Work out 4 x 15 7. Give your answer as a mixed number.

Mathematics
1 answer:
hjlf3 years ago
8 0

Answer:

60/7

Step-by-step explanation:

4/1 x 15/7 =8 4/7 = 60/7

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Evaluate log38, given log32 ≈ 0.631.
ArbitrLikvidat [17]
\log_38-\log_32^3=3\log_32\approx3\cdot0.631=1.893\\\\Used:\log_ab^n=n\cdot\log_ab
4 0
3 years ago
Read 2 more answers
A study was done to determine the average number of homes that a homeowner owns in his or her lifetime. For the 60 homeowners su
const2013 [10]

Answer:

D) (3.67, 4.73)

Step-by-step explanation:

Confidence Interval for the true average number of homes that a person owns in his or her lifetimecan be calculated using M±ME where

  • M is the average number of home owned (4.2)
  • ME is the margin of error from the mean

And margin of error (ME) can be calculated as

ME=\frac{z*s}{\sqrt{N} } where

  • z is the corresponding statistic in the given confidence level(1.96)
  • s is the standard deviation of the sample(2.1)
  • N is the sample size (60)

Putting the numbers we get ME=\frac{1.96*2.1}{\sqrt{60} }≈0.53

Then the 95% confidence interval is 4.2±0.53 or (3.67, 4.73)

6 0
3 years ago
Solve for f. d(e-f)=g
Yakvenalex [24]

Isolate the variable by dividing each side by factors that don't contain the variable.

f=e-g/d

8 0
3 years ago
Solve 3cot (2x) =- Route3 for<br> x, 0 = x &lt; 360
ivolga24 [154]

Answer:

x=60,150,240,330

Step-by-step explanation:

3 \cot(2x)  =  -  \sqrt{3}

\cot(2x)  =  \frac{  - \sqrt{3} }{3}

Take the arc cot of both sides

\cot {}^{ - 1} ( \cot(2x) )  =  \cot {}^{ - 1} (  \frac{ -  \sqrt{3} }{3}  )

2x = 120

Remember cotangent has a period of 180 degrees

2x = 120 + 90(n)

where n is 0, 1,2,3,4,5.....

Isolate x.

x = 60 + 90(n)

where n is 0, 1,2,3,4,5,6.

Keep plugging in integers as long they are in the interval [0,360].

We get

60,150,240,330.

6 0
2 years ago
Determine equation for line that is perpendicular to x=-2 and passes through the point (6,1)​
Zina [86]

Answer:

y = 1

Step-by-step explanation:

6 0
3 years ago
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