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GREYUIT [131]
3 years ago
10

Please i need help and this is due in an hour

Mathematics
2 answers:
joja [24]3 years ago
7 0

Answer:

x=11.2

Step-by-step explanation:

cross multiple 100 over x equals 200 over 22.4

100(22.4) = 200x

2,240 = 200x

divide

11.2=x

Ksju [112]3 years ago
3 0

Answer:

\frac{100 meters}{x seconds}  = \frac{200 meters}{22.4 seconds}

Step-by-step explanation:

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FRAME AN EQUATION AND SOLVE IT. NEED HELP ASAP I only have 5 minutes!​
lawyer [7]

Answer:

<em>The answer to your question is</em><em> 125</em>

Step-by-step explanation:

<em> x = 25 X 5= 125.</em>

<em>The required number is 125.</em>

<u><em>I hope this helps and have a good day!</em></u>

3 0
2 years ago
00:00
rusak2 [61]

Answer:

I thinks no for all except the last one.

6 0
2 years ago
Find the sum and express it in simplest form. (ab+4a-6)+(ab+6)
riadik2000 [5.3K]
= 2ab + 4a + 6 - 6

= 2ab + 4a

you can factor this  to 2a(b + 2)
3 0
3 years ago
Read 2 more answers
Express 15 kobo as a decimal of #3.00
arsen [322]

Answer:

15 kobocoin =30.143471

5 0
2 years ago
A company claims that the mean weight per apple they ship is 120 grams with a standard deviation of 12 grams. Data generated fro
Veronika [31]

Answer:

There is no sufficient evidence to reject the company's claim at the significance level of 0.05

Step-by-step explanation:

Let \mu be the true mean weight per apple the company ship.  We want to test the next hypothesis

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Because we have a large sample of size n = 49 apples randomly selected from a shipment, the test statistic is given by

Z=\frac{\bar{X}-120}{\sigma/\sqrt{n}} which is normally distributed. The observed value is  

z_{0}=\frac{122.5-120}{12/\sqrt{49}}=1.4583. The rejection region for \alpha = 0.05 is given by RR = {z| z < -1.96 or z > 1.96} where the area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well. Because the observed value 1.4583 does not fall inside the rejection region RR, we fail to reject the null hypothesis.

5 0
3 years ago
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