In this case we know the three sides of the triangle, then this is a SSS triangle (Side Side Side). To solve this case, first we must use the Law of Cosines, applied to the opposite side to the angle we want to find.
We want to find angle W, and its opposite side is XV, then we apply the Law of Cosines to the side XV:
XV^2=XW^2+WV^2-2(XW)(WV)cos W
Replacing the known values:
116^2=96^2+89^2-2(96)(89)cos W
Solving for W
13,456=9,216+7,921-17,088 cos W
13,456=17,137-17,088 cos W
13,456-17,137=17,137-17,088 cos W-17,137
-3,681=-17,088 cos W
(-3,681)/(-17,088)=(-17,088 cos W)/(-17,088)
0.215414326=cos W
cos W = 0.215414326
Solving for W:
W= cos^(-1) 0.215414326
Using the calculator:
W=77.56016397°
Rounded to one decimal place:
W=77.6°
Answer: Third option 77.6°
<span>it is the letter b because if you count closly you will see
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Answer:
the 24-ounce box
Step-by-step explanation:
18/2.88 = 6.25$
24/3.6 = 6.67$
Answer:
20) 2/32 so colour 2 boxes
21) 7/16
22) 2/60
23) 3/8
24) 35/63 or 5/9
25) 7/32
26) 45/60 or 9/12 or 3/4
27) 1/32
28) 3/21 or 1/7
29) 13/6 × 9/2
= 117/12
30) 3/4 × 17/2
= 51/8
31) 9/8 × 10/3
= 90/24
= 45/12
= 15/4
32) 16/5 × 2/3
= 32/15
sorry if theres any calculation mistake becoz i did all the calculations in mind
Step-by-step explanation:
Answer:
a)
b) 
c) 
d) 
And the dsitribution that satisfy this is 
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Let X the random variable of interest, on this case we now that:

The probability mass function for the Binomial distribution is given as:

Where (nCx) means combinatory and it's given by this formula:

Part a
Part b
The expected value is givn by:

Part c
For the standard deviation we have this:

Part d
For this case the sample size needs to be higher or equal to 9. Since we need a value such that:

And the dsitribution that satisfy this is 
We can verify this using the following code:
"=1-BINOM.DIST(3,9,0.75,TRUE)" and we got 0.99 and the condition is satisfied.