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Vinvika [58]
3 years ago
6

What is the value of 5 2/3(−3.6)?

Mathematics
1 answer:
Anestetic [448]3 years ago
4 0

Answer:

<em>Hello, Your answer will be -20 2/4.</em>

Step-by-step explanation:

Conversion a mixed number 5 2

3

to a improper fraction: 5 2/3 = 5 2

3

= 5 · 3 + 2

3

= 15 + 2

3

= 17

3

To find new numerator:

a) Multiply the whole number 5 by the denominator 3. Whole number 5 equally 5 * 3

3

= 15

3

b) Add the answer from previous step 15 to the numerator 2. New numerator is 15 + 2 = 17

c) Write previous answer (new numerator 17) over the denominator 3.

Five and two thirds is seventeen thirds

Conversion a decimal number to a fraction: -3.6 = -36

10

= -18

5

a) Write down the decimal -3.6 divided by 1: -3.6 = -3.6

1

b) Multiply both top and bottom by 10 for every number after the decimal point. (For example, if there are two numbers after the decimal point, then use 100, if there are three then use 1000, etc.)

-3.6

1

= -36

10

Note: -36

10

is called a decimal fraction.

c) Simplify and reduce the fraction

-36

10

= -18 * 2

5 * 2

= -18 * 2

5 * 2

= -18

5

Multiple: 17

3

* (-3.6) = 17 · (-18)

3 · 5

= -306

15

= -102 · 3

5 · 3

= -102

5

Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(-306, 15) = 3. In the next intermediate step the fraction result cannot be further simplified by cancelling.

In words - seventeen thirds multiplied by minus eighteen fifths = minus one hundred two fifths. <em>Hope That Helps!</em>

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Answer:

a-1) Reject H0 if zcalc > 1.645

a-2) z=\frac{0.316 -0.28}{\sqrt{\frac{0.28(1-0.28)}{114}}}=0.8561  

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Step-by-step explanation:

1) Data given and notation

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p_o=0.28 is the value that we want to test

\alpha=0.05 represent the significance level

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We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.28.:  

Null hypothesis:p\geq 0.28  

Alternative hypothesis:p < 0.28  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

The rejection zone would be on this case :

Reject H0 if zcalc > 1.645

Since is a right tailed test

(a-2) Calculate the test statistic. (Round intermediate calculations to 2 decimal places. Round your answer to 4 decimal places.) Test statistic

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.316 -0.28}{\sqrt{\frac{0.28(1-0.28)}{114}}}=0.8561  

(a-3) The null hypothesis should be rejected.

False, since our calculated value is less than our critical value we Fails to reject the null hypothesis

(b) Is this a close decision?

False the calculated value is significantly less than the critical value so we FAIL to reject the null hypothesis with enough confidence.

(c) State any assumptions that are required.

In order to satisfy the conditions we need the following two requirements:

iii. n π > 10 and n(1 − π ) > 10

And are satisfied:

114*0.28=31.92>10

114(1-0.28)=82,08>10

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