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Diano4ka-milaya [45]
2 years ago
8

8. A map has a scale factor of 0.25 inch : 50 miles. On the map, the distance of a river is 2.5 inches. What is the

Mathematics
1 answer:
Arturiano [62]2 years ago
7 0

Set up a ratio for this problem:

.25 in/50 miles = 2.5 in/x

.25x = 125

then divide by .25

The actual distance is: 500 miles

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Which expression is equivalent to this expression? x(2x+3) Question 1 options: 2x2+3 2x2+3x 2x2−3x 3x + 3
sleet_krkn [62]

Answer:

2x^2+3x is the expression which is equivalent to  x\left(2x+3\right).

In other words:

x\left(2x+3\right)=2x^2+3x

Step-by-step explanation:

Given the expression

x\left(2x+3\right)

solving

x\left(2x+3\right)

\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac

a=x,\:b=2x,\:c=3

=x\cdot \:2x+x\cdot \:3

=2xx+3x

=2x^2+3x            ∴  2xx=2x^2

Thus,

x\left(2x+3\right)=2x^2+3x

Therefore, 2x^2+3x is the expression which is equivalent to  x\left(2x+3\right).

8 0
3 years ago
The LB Company has long manufactured a light bulb with an average life of 5400 hours. Company researchers have recently develope
elena55 [62]
<h2><u>Answer with explanation</u>:</h2>

Let \mu be the average life of light bulbs.

As per given , we have

Null hypothesis : H_0 : \mu =5400

Alternative hypothesis : H_a : \mu >5400

Since H_a is right-tailed and population standard deviation is also known, so we perform right-tailed z-test.

Formula for Test statistic : z=\dfrac{\overlien{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

where, n= sample size

\overline{x}= sample mean

\mu= Population mean

\sigma=population standard deviation

For n=95,\ \overline{x}=5483\ \&\ \sigma=500, we have

z=\dfrac{5483-5400}{\dfrac{500}{\sqrt{95}}}=1.61796786124\approx1.6180

Using z-value table , Critical one-tailed test value for 0.06 significance level :

z_{0.06}=1.5548

 Decision : Since critical z value (1.5548) < test statistic (1.6180), so we reject the null hypothesis .

[We reject the null hypothesis when critical value is less than the test statistic value .]

Conclusion : We have enough evidence at 0.06 significance level to support the claim that the new filament yields a longer bulb life

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