What is the measure of XYZ 36 Degrees Please help

The animal shelter has both male and female Labrador Retrievers in yellow, brown, or black. There is an equal number of each kin

castortr0y [4]

school is asking us to find the probability of dogs breeding I’m sorry I don’t understand

7 0

3 years ago

Will give brainlst!!

Sedaia [141]

2.5
,
21
8
,
3
√
25
,
π
,
√
19
2.5
,
21
8
,
25
3
,
π
,
19

3 0

3 years ago

A circle is drawn inside a square.

zhannawk [14.2K]

Shaded area = 21.46

Step-by-step explanation:

First find area of square,

Area = length × width

A = 20 × 20

A = 400 cm^2

Secondly, find area of circle

Area of circle = (π)r^2

A = (π)10^2

A = 100 π

Thirdly, find the area outside of the circle

Area of square - Area of circle

400cm^2 - 100π

= 85.84

lastly divide the answer by 4 to find the shaded area

85.84 ÷ 4

= 21.46

5 0

2 years ago

G fewer than the quotient of h and f

dangina [55]

Answer:

h/f - g

Step-by-step explanation:

Write "the quotient of h and f" first: h/f

"g fewer than this quotient" is expressed as h/f - g.

4 0

3 years ago

A college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of

Alla [95]

Answer:

80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

Step-by-step explanation:

We are given that a college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of hours they watch per week. The results are provided below;

Hours of TV per week (X): 6, 14, 13, 6, 16, 10, 19, 4, 5, 5, 18, 8, 7, 14, 8, 8, 9, 12, 6, 5.

Firstly, the Pivotal quantity for 80% confidence interval for the true average is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean number of hours of TV watched per week = $\frac{\sum X}{n}$ = 9.65

s = sample standard deviation = $\sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }$ = 4.61

n = sample of people = 20

$\mu$ = true average number of hours of TV watched per week

Here for constructing 80% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the true average, $\mu$ is ;

P(-1.33 < $t_1_9$ < 1.33) = 0.80 {As the critical value of t at 19 degrees of

freedom are -1.33 & 1.33 with P = 10%}

P(-1.33 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.33) = 0.80

P( $-1.33 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $9.65-1.33 \times {\frac{4.61}{\sqrt{20} } }$ , $9.65+1.33 \times {\frac{4.61}{\sqrt{20} } }$ ]

= [8.28 hours, 11.02 hours]

Therefore, 80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

7 0

3 years ago