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Mrrafil [7]
3 years ago
6

it took 8 hours for karl and lou to finish a project for science class together. if karl had worked alone on the project it woul

d have taken him 10 hours. how long would it have taken lou to complete the project by himself.
Mathematics
1 answer:
kkurt [141]3 years ago
4 0
It would take 10 hours just like it did for Karl .
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Solve the equation.
blondinia [14]

Answer: 12

Step-by-step explanation:

6c + 11= 2c + 59

Collect like terms

6c - 2c = 59 - 11

4c = 48

c = 48/4

c = 12

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3 years ago
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What's the value of given expression :<br><br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B9%7D%20" id="TexFormula1" title=" \
shusha [124]

Answer:

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2 years ago
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Need help on this one
malfutka [58]

Answer:

3

Step-by-step explanation:

To simplify this expression, we will simply work with the numerator, then the denominator, and then the fraction as a whole, following PEMDAS.

Numerator:  12 - 6 + 5 * (-3)^2

= 12 - 6 + 5 * (9)

= 12 - 6 + 45

= 6 + 45

= 51

Denominator:  17

So our simplified fraction is as follows:

51/17

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Cheers.

3 0
3 years ago
A rectangular lawn is twice as long as its breadth and its perimeter is 90 m.
Nina [5.8K]
Answer: length = 30, breadth = 15

length = l
breadth = b

l = 2b

2b + b + 2b + b = 90
6b = 90
b = 15

l = 2b
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length = 30

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7 0
3 years ago
Using traditional methods, it takes 95 hours to receive a basic flying license. A new license training method using Computer Aid
Snowcat [4.5K]

Answer:

There is not sufficient evidence to support the claim that the technique performs differently than the traditional method.

Step-by-step explanation:

The null hypothesis is:

H_{0} = 95

The alternate hypotesis is:

H_{1} \neq 95

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

A researcher used the technique with 260 students and observed that they had a mean of 94 hours. Assume the standard deviation is known to be 6.

This means, respectively, that n = 260, X = 94, \sigma = 6

The test-statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{94 - 95}{\frac{6}{\sqrt{260}}}

z = -2.69

The pvalue is:

2(P(Z < -2.69))

P(Z < -2.69) is the pvalue of Z when X = -2.69, which looking at the z-table, is 0.0036

2*(0.0036) = 0.0072

0.0072 < 0.01, which means that the null hypothesis is accepted, that is, there is not sufficient evidence to support the claim that the technique performs differently than the traditional method.

7 0
2 years ago
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