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3 years ago

Find the value of x........

Mathematics

1 answer:

raketka [301]3 years ago

7 0

Answer:

x=23

Step-by-step explanation:

Hello There!

Remember the exterior angle of a triangle is equal to the opposite interior angles of a triangle

137-x=2x+3x-1

now we can solve for x

step 1 combine like terms

2x+3x=5x

now we have

137-x=5x-1

step 2 add 1 to each side

-1+1 cancels out

137+1=138

138-x=5x

step 2 add x to each side

-x+x cancels out

5x+x=6x

now we have

138=6x

step 3 divide each side by 6

6x/6=x

138/6=23

we´re left with x=23

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3 years ago

A market research company conducted a survey to find the level of affluence in a city. They defined the category "affluence" for

malfutka [58]

Answer:

A 95% confidence interval for this population proportion is [0.081, 0.159].

Step-by-step explanation:

We are given that a market research company conducted a survey to find the level of affluence in a city.

Out of 267 persons who replied to their survey, 32 are considered affluent.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of people who are considered affluent = $\frac{32}{267}$ = 0.12

n = sample of persons = 267

p = population proportion

Here for constructing a 95% confidence interval we have used One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.12-1.96 \times {\sqrt{\frac{0.12(1-0.12)}{267} } }$ , $0.12+1.96 \times {\sqrt{\frac{0.12(1-0.12)}{267} } }$ ]

= [0.081, 0.159]

Therefore, a 95% confidence interval for this population proportion is [0.081, 0.159].

4 0

3 years ago