A study was conducted by a research center. It reported that most shoppers have a specific spending limit in place while shopping online. The reports indicate that men spend an average of $240 online before they decide to visit a store. If the spending limit is normally distributed and the standard deviation is $20. A. Find the probability that a male spent less than $210 online before deciding to visit a store.
B. Find the probability that a male spent between $270 and $300 online before deciding to visit a store.
C. Ninety percent of the amounts spent online by a male before deciding to visit a store are less than what value?

(A) The probability that a male spent less than $210 online before deciding to visit a store is 0.0668.

(B) The probability that a male spent between $270 and $300 online before deciding to visit a store is 0.0655.

(C) Ninety percent of the amounts spent online by a male before deciding to visit a store is less than $265.632.

Step-by-step explanation:

We are given that the reports indicate that men spend an average of $240 online before they decide to visit a store. If the spending limit is normally distributed and the standard deviation is $20.

Let X = the spending limit

The z-score probability distribution for the normal distribution is given by;

                           Z  =    ~ N(0,1)

where, = mean spending limit = $240

            = standard deviation = $20

So, X ~ Normal()

(A) The probability that a male spent less than $210 online before deciding to visit a store is given by = P(X < $210)

     P(X < $210) = P( < ) = P(Z < -1.50) = 1 - P(Z 1.50)

                                                            = 1 - 0.9332 = 0.0668

The above probability is calculated by looking at the value of x = 1.50 in the z table which has an area of 0.9332.

(B) The probability that a male spent between $270 and $300 online before deciding to visit a store is given by = P($270 < X < $300)

     P($270 < X < $300) = P(X < $300) - P(X $270)

     P(X < $300) = P( < ) = P(Z < 3) = 0.9987

     P(X $270) = P( ) = P(Z 1.50) = 0.9332

The above probability is calculated by looking at the value of x = 3 and x = 1.50 in the z table which has an area of 0.9987 and 0.9332 respectively.

Therefore, P($270 < X < $300) = 0.9987 - 0.9332 = 0.0655.

(C) Now, we have to find ninety percent of the amounts spent online by a male before deciding to visit a store is less than what value, that is;

         P(X < x) = 0.90     {where x is the required value}

         P( < ) = 0.90

         P(Z < ) = 0.90

In the z table, the critical value of z that represents the bottom 90% of the area is given as 1.2816, i.e;

                     x = 265.632

Hence, Ninety percent of the amounts spent online by a male before deciding to visit a store is less than $265.632.

b

step-by-step explanation:

u take $75 minus $60.the anwer/original($75)multiply by 100

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)