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My name is Ann [436]
3 years ago
13

melinda has a map of her city. the map uses a scale of 1 inch = 8 mile. Melinda's house is 1 1/2 inches away from the library on

the map. how far apart would her house and the library be on the map if the scale were 1 inch = 6 mile?
Mathematics
1 answer:
BigorU [14]3 years ago
3 0

Answer:

2 inches

Step-by-step explanation:

her house is 12 miles away from the library so if its 1 inch per 6 mile

then multiply both by 2 to get 2 inches for 12 miles

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Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab
Katyanochek1 [597]

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected,  and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, H_0 : \mu = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, H_1 : \mu < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

       T.S. = \frac{Xbar-\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, X bar = sample mean = 738.44 hours

               s  = sample standard deviation = 38.2 hours

               n = sample size = 50

So, test statistics = \frac{738.44-750}{\frac{38.2}{\sqrt{50} } } ~ t_4_9

                            = -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

6 0
3 years ago
Suppose for some value of x the solution to the equation 2.5(y−x)=0 is y = 6. What must be true about x?
Varvara68 [4.7K]

Step-by-step explanation:

In this case, you input the value of y (y = 6) into the equation ( 2.5(y-x)= 0)

2.5(6-x) = 0

Open the bracket,

(2.5×6)-2.5x= 0

Collect like terms.

2.5x= 2.5×6

Divide both sides by the coefficient of x (2.5)

x = 6

So,

x= 6 is true.

.

Substituting 6 for x in the equation,

2.5(6-6)= 0

2.5•0 = 0

0= 0

Which is also true.

7 0
2 years ago
Need help on both of them
Vlad1618 [11]
7) Write the decimal number as a fraction
(over 1)
2.18 = 2.18 / 1

Multiplying by 1 to eliminate 3 decimal places
we multiply top and bottom by 3 10's

Numerator (N)
N = 2.18 × 10 × 10 × 10 = 2180
Denominator (D)
D = 1 × 10 × 10 × 10 = 1000

N / D = 2180 / 1000

Simplifying our fraction

= 2180/1000

= 109/50

= 2  9/50

8)7 ^{-5}  =  \frac{1}{7^{5}}
6 0
3 years ago
In terms of the number of moose, what is the relative frequency of male moose, female moose, adult moose, and baby moose? Write
user100 [1]

Answer:

I'm sorry is there a chart/graph, etc. to give us more information?

Without the added information there is no way for me to answer.

If you repost this question with a picture I would be more than happy to help.

6 0
3 years ago
Read 2 more answers
Can someone please help me with this!?!?!
lys-0071 [83]
They are congruent because of the angles, which means the congruency reason should be SAS.
4 0
3 years ago
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