Answer:
9.5 feets
Step-by-step explanation:
Length of drawing = 8 4/9 inches
Scale : 1/3 inch in drawing = 3/8 actual length
Length of drawing / drawing scale
8 4/9 = 76/9 inches
Number of 1/3 inches in total length of drawing :
76/9 ÷ 1/3
76/9 * 3 / 1
(76*3) /(9*1) = 228 /9 = 9 3/9 = 9 1/3
9 1/3 * 3/8 foot
(228 / 9) * (3 /8)
(228/3) * (1/ 8)
228 / 24
= 9.5 feet
10000 digits can be used for 4 digit A.T.M code.
<u>Solution:</u>
Given that A.T.M required 4 digit codes using the digits 0 to 9.
Need to determine how many four digit code can be used.
We are assuming that number starting with 0 are also valid ATM codes that means 0789 , 0089 , 0006 and 0000 are also valid A.T.M codes.
Now we have four places to be filled by 0 to 9 that is 10 numbers
Also need to keep in mind that repetition is allowed in this case means if 9 is selected at thousands place than also it is available for hundreds, ones or tens place .
First digit can be selected in 10 ways that is from 0 to 9.
After selecting first digit, second digit can be selected in 10 ways that is 0 to 9 and same holds true for third and fourth digit.
So number of ways in which four digit number is created = 10 x 10 x 10 x 10 = 10000 ways
Hence 10000 digits can be used for 4 digit A.T.M code.
Answer:
true
Step-by-step explanation:
a square will always be a rectangle but a rectangle will not always be a square.
Complete question :
Suppose that of the 300 seniors who graduated from Schwarzchild High School last spring, some have jobs, some are attending college, and some are doing both. The following Venn diagram shows the number of graduates in each category. What is the probability that a randomly selected graduate has a job if he or she is attending college? Give your answer as a decimal precise to two decimal places.
What is the probability that a randomly selected graduate attends college if he or she has a job? Give your answer as a decimal precise to two decimal places.
Answer:
0.56 ; 0.60
Step-by-step explanation:
From The attached Venn diagram :
C = attend college ; J = has a job
P(C) = (35+45)/300 = 80/300 = 8/30
P(J) = (30+45)/300 = 75/300 = 0.25
P(C n J) = 45 /300 = 0.15
1.)
P(J | C) = P(C n J) / P(C)
P(J | C) = 0.15 / (8/30)
P(J | C) = 0.5625 = 0.56
2.)
P(C | J) = P(C n J) / P(J)
P(C | J) = 0.15 / (0.25)
P(C | J) = 0.6 = 0.60
Answer:
The player with the most runs had a rush of 1,240 yards
Step-by-step explanation:
In this question, we are asked to calculate the number of yards that was rushed by one of two person given their combined run and an extra information.
Firstly, let the person that had the smaller number of rush have a rush of x rushes. The second person has a rush of 4 times the other. This makes a number of 4x rushes
By adding both together, we have a total of 1550 yards
Mathematically, this means that x + 4x = 1550
5x = 1550
x = 1550/5 = 310 rushes
The second player had a rush of 4x and that is 4 * 310 = 1,240 rushes
