Pls help i will give the brielliast answer

Madhu can select from 3 type of oranges and 4 types apples. If she randomly selei 1 orange and 1 apple,how many possible choices

viktelen [127]

(1 orange/3 possibilities) * (1 apple/4 possibilities) = (1 result/12 possible results)
She has 12 possible choices

8 0

3 years ago

Social Networking Sites

SOVA2 [1]

Answer:

(a). ( 0.776 ,0.809).

(b). (0.567 , 0.613).

(c). 0.600.

Step-by-step explanation:

Okay, we are given the following set of values or data or parameters;

=> "A survey of 2255 randomly selected US adults (age 18 or older)"

=> "1787 of them use the Internet regularly. Of the Internet users, 1054 use a social networking site".

=> Also, "95% confidence interval for each of the following proportions"

Therefore, we are going to make use of one (major ) mathematical formula in solving this particular Question and it is given below;

Confidence Interval = p +/- z* × [ √p( 1 - p) / n].

(a).

Where p = 1787/2255 = 0.793.

95% confidence Interval = z* = 1.96.

= 0.793 +/- 1.96 × [√0.793 ( 1 - 0.793)/ 2255] .

= 0.793 +/- 0.0167.

= ( 0.776 ,0.809).

(b). Where p = 1054/ 1787 = 0.5900

95% confidence Interval = z* = 1.96.

= 0.5900 +/- 1.96 × [√0.5900 ( 1 - 0.5900)/ 1787]

= 0.5900 +/- 0.0228.

= (0.567 , 0.613).

(c). 1054/1787 = 0.59 = 0.600.

3 0

3 years ago

Brainliest I will give pls help me

Angelina_Jolie [31]

Step-by-step explanation:

a) complimentary angles = 90°

so,

let the complement be x, 

so after inserting values we got,

72.5° + x = 90°

→ x = 90.0 - 72.5

→ x = 17.5° ans

supplementary angles = 180°

let the supplement of 72.5° be x

so, after inserting values we got,

72.5 + x = 180°

→ x = 180.0 - 72.5

→ x = 107.5° ans

and I have no idea about b...but hope this answer helps you dear!

5 0

3 years ago

Can someone help me with this please

MArishka [77]

(x-5)(x+4)

The two numbers should add up to the b value and multiply to the c value.
Check:
-5+4=-1
-5*4=-20

Final answer: B

7 0

3 years ago

Hereford cattle are one of the most popular breeds of beef cattle. Based on data from the Hereford Cattle Society, the mean weig

loris [4]

Let x be a random variable representing the mean weight of a one-year-old Hereford bull. Then
a.) P(x < 1100) = P (z < (X - mean)/sd) = P(z < (1100 - 1150)/80) = P(z < -0.625) = 1 - P(z < 0.625) = 1 - 0.73401 = 0.2660

b.) The middle (50%) of Hereford weights will weigh the mean weight of the sample. i.e. the middle (50%) of the Hereford weights will be 1150 pounds

5 0

4 years ago