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kap26 [50]
3 years ago
11

In the diagram below segment DEF is shown (since all three letters are shown, they are collinear). If

Mathematics
1 answer:
monitta3 years ago
3 0

Answer:

  • x = 4

Step-by-step explanation:

<u>Given</u>

  • DE = 3x + 2, EF = x + 5, DF = 23

<u>As per segment addition postulate</u>

  • DE + EF = DF

<u>Substituting values:</u>

  • 3x + 2 + x + 5 = 23
  • 4x + 7 = 23
  • 4x = 16
  • x = 4
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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
2 years ago
HELP PLEASE!!! I need help with this question!
AlexFokin [52]

Answer:

\large\boxed{PS=10\sqrt5}

Step-by-step explanation:

ΔPQS, ΔRQP and ΔRPS are similar (AA). Therefore the sides are in proportion:

\dfrac{QR}{RP}=\dfrac{RP}{RS}

We have:

QR=5,\ RS=20

Substitute:

\dfrac{5}{RP}=\dfrc{RP}{20}                <em>cross multiply</em>

RP^2=(5)(20)\\\\RP^2=100\to RP=\sqrt{100}\\\\RP=10

Use the Pythagorean theorem:

PS^2=PR^2+RS^2

Substitute:

PS^2=10^2+20^2\\\\PS^2=100+400\\\\PS^2=500\to PS=\sqrt{500}\\\\PS=\sqrt{100\cdot5}\\\\PS=\sqrt{100}\cdot\sqrt5\\\\PS=10\sqrt5

6 0
2 years ago
Simplify this algbraic expression completely 5y-3(y+2)
Sophie [7]

Answer:

2y - 6

Step-by-step explanation:

5y - 3(y + 2)

distribute

5y + (-3 * y) + (-3 * 2)

simplify

5y -3y + ( -3 * 2)

5y - 3y + ( -6)

5y - 3y - 6

combine like terms

5-3 = 2

2y - 6

3 0
2 years ago
Find the proportion of observations from a standard Normal distribution that falls in each of the following regions. -2.31 &lt;
Igoryamba
\mathbb P(-2.31
4 0
2 years ago
Bella measured these outside temperatures for each of four days.
kap26 [50]
What is the complete question and what are the choices??
4 0
3 years ago
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