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2 years ago

Convert 14.75 to its equivalent fraction expressed in lowest terms

Mathematics

1 answer:

stellarik [79]2 years ago

6 0

Either its 14 and 3 4ths or 59 4ths. its depends on what fractions it wants it in. hope this makes sense!

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? A survey of factories in five northeastern states found that 10% of the 300 workers surveyed were satisfied with the benefits

Anastasy [175]

Percent means per one-hundred

300(10/100)=30

So 30 of the 300 workers were satisfied with their benefits.

5 0

2 years ago

Look at the figure. Which of the following segments is parallel to AB?

marusya05 [52]

I think it’s c and h

5 0

2 years ago

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find

fenix001 [56]

Answer:

y2 = (6x + 7)/36 + (Dx + E)e^x

Step-by-step explanation:

The method of reduction of order is applicable for second-order differential equations.

For a known solution y1 of a 2nd order differential equation, this method assumes a second solution in the form Uy1 which satisfies the said differential equation. It then assumes a reduced order for U'' (w' = U'').

The differential equation becomes easy to solve, and all that is left are integration and substitutions.

Check attachments for the solution to this problem.

6 0

1 year ago

g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the

blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean amount of money spent on textbooks = $121.60

$\sigma$ = population standard deviation = $6.36

n = sample of students = 36

$\mu$ = population mean

Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $121.60-1.645 \times {\frac{6.36}{\sqrt{36} } }$ , $121.60+1.645 \times {\frac{6.36}{\sqrt{36} } }$ ]

= [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0

2 years ago

Solve 9/11 + 3/13 × 26/33 - (21/121÷ 71/111)

Julli [10]

Answer:

0.72 or 0.72866953788 0.7 0.72866953788

3 0

2 years ago