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GaryK [48]
3 years ago
12

Please help I'll give you brainliest

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
7 0

Answer:

Equation 1: No solution

Equation 2: Infinity

Equation 3: One solution

Equation 4: Infinity

Equation 5: No solution

Hope this helps you. Do mark my answer as brainly. Thanks

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disa [49]

Answer:

I believe your answer is correct.

Step-by-step explanation:

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2 years ago
Bruno is 4.5 times as old as Charles if Bruno is 27 how old is Charles
andrey2020 [161]

Answer:

6

Step-by-step explanation:

Bruno(B)= 4.5× Charles(C)

8 0
2 years ago
Read 2 more answers
Hello can you please help me posted picture of question
zhuklara [117]
Since the order of selection does not matter, we will use combinations to solve this problem.

We are to form the combination of 30 objects taken 6 at a time. This can be expressed as 30C6.

30C6= \frac{30!}{6!*24!}  \\  \\ 
= 593775

This means the six teachers can be selected in 593775 ways.

So the correct answer is option A
3 0
3 years ago
Suppose y varies directly with x. When x is 2, y is 20. What is x when y is 50
yuradex [85]

Answer:

5

Step-by-step explanation:

x = 2 >>> (Multiply by 10) >>> y = 20

x = ? >>> (Multiply by 10) >>> y = 50

50 ÷ 10 = 5

x = 5 when y is 50

5 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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