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sergeinik [125]
2 years ago
8

How much longer is 18.25 centimeters than 75.4 milimeters

Mathematics
1 answer:
Amanda [17]2 years ago
4 0

Answer:

10.71

Step-by-step explanation:

You first convert 75.4 millimeters to centimeters, and you will get 7.54. Now you subtract 18.25 and 7.54 to get the answer 10.71

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2 3/5 convert to decinal
Leokris [45]

So we know that 1/5=0.2

Then we take 5/5=1.0

Then we solve 2*1.0=2.0

Now we solve 1/5+1/5+1/5 or 1/5*3=0.6

So then the correct answer is 2.6

Hope this helps!!!

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7 0
3 years ago
Im sorry it’s blurry! can someone please help me out it would mean a lot thank you!
alekssr [168]

Answer: B. -3, -2, 1, 2, 5

Step-by-step explanation:

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3 0
2 years ago
The cost of a cycle is $ 950 and that of a scooter is $ 23,500. He sold them together for $ 25,000. Find his profit. ​
e-lub [12.9K]

actual cost is 950+23,500=24450

he sold them for 25000 so his profit is 25000-950-23500=550

4 0
3 years ago
Read 2 more answers
A vector is in standard position, with its terminal point in the second quadrant and an x-coordinate of -5. The vector
Nonamiya [84]

Answer:6.0, 130

Step-by-step explanation:

Just got the question right on edgenuity.

3 0
3 years ago
A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.
TEA [102]

Answer:

(a) Null Hypothesis, H_0 : \mu = $1,150  

    Alternate Hypothesis, H_A : \mu \neq $1,150

(b) The test statistic is 3.571.

(c) We conclude that the mean of all account balances is significantly different from $1,150.

Step-by-step explanation:

We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.

<u><em>Let </em></u>\mu<u><em> = mean of all account balances</em></u>

(a)So, Null Hypothesis, H_0 : \mu = $1,150     {means that the mean of all account balances is equal to $1,150}

Alternate Hypothesis, H_A : \mu \neq $1,150    {means that the mean of all account balances is significantly different from $1,150}

The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about population standard deviation;

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

where, \bar X = sample average balance = $1,200

             s = sample standard deviation = $126

             n = sample of account balances = 81

(b) So, <u>test statistics</u>  =  \frac{1,200-1,150}{\frac{126}{\sqrt{81} } }  ~ t_8_0

                               =  3.571

The value of the sample test statistics is 3.571.

(c) <u>Now, P-value of the test statistics is given by the following formula;</u>

          P-value = P( t_8_0 > 3.571) = <u>Less than 0.05%</u>

Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.

Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean of all account balances is significantly different from $1,150.

8 0
3 years ago
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