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Ymorist [56]
3 years ago
6

Find two consecutive odd integers such that 53 more than the lesser is four times the greater

Mathematics
1 answer:
viva [34]3 years ago
4 0
Let x and x+2 the 2 odd consecutive numbers. Now translate from English language to Math's:
x+53 = 4(x + 2), Expand :
x+53 = 4x + 8
53 - 8 = 4x - x

45 = 3x   and x = 15. So the 1st is 15 and the 2nd is 17

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What is the answer to (t+8)(-2)=12
Lunna [17]
(t + 8) (-2) = 12

Distributive property >>>      -2t + -16 = 12

Inverse operation >>>>    -2t + -16+16 = 12+16
 
Simplify >>>>                 -2t = 28

Divide >>>>                       -2t/-2 = 28/-2

Final answer >>>>              t = -14
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I think it's 300

Step-by-step explanation:

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What is the propibility of getting a pizza with plain crust
Katen [24]

Answer:

50%

Step-by-step explanation:

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PLS HELP!!!! ASAP
Umnica [9.8K]

Answer:

PLS HELP!!!! ASAP

Like terms 3x, 2x

Different terms 4y, 7z

Step-by-step explanation:

Example similar terms

3x + x = 5x

Example different terms

4 years + 7z

4 0
3 years ago
Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o
kirza4 [7]

Answer:

a) H0: \sigma = 1.34

H1: \sigma \neq 1.34

b) df = n-1= 10-1=9

And the critical values with \alpha/2=0.005 on each tail are:

\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589

c) t=(10-1) [\frac{1.186}{1.34}]^2 =7.05

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

\sigma_o =1.34 the value that we want to test

p_v represent the p value for the test

t represent the statistic  (chi square test)

\alpha=0.01 significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: \sigma = 1.34

H1: \sigma \neq 1.34

The statistic is given by:

t=(n-1) [\frac{s}{\sigma_o}]^2

Part b

The degrees of freedom are given by:

df = n-1= 10-1=9

And the critical values with \alpha/2=0.005 on each tail are:

\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589

Part c

Replacing the info we got:

t=(10-1) [\frac{1.186}{1.34}]^2 =7.05

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0
3 years ago
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