bearing in mind that there are 90 feet between each base in a baseball diamond, so the total perimeter of it will be 90*4 = 360, one third of that is 360/3 = 120.
So Jones went 120 feet first and then another and so on, check the picture below.
using those values, can get the length of those sides, using the pythagorean theorem.
![\bf AH=\sqrt{HF^2+FA^2}\implies AH=\sqrt{9000} \\\\\\ AB=\sqrt{AS^2+SB^2}\implies AB=\sqrt{7200} \\\\\\ HB = AH = \sqrt{9000} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{perimeter}{\sqrt{9000}+\sqrt{7200}+\sqrt{7200}}\qquad \approx \qquad \stackrel{perimeter}{264.57}](https://tex.z-dn.net/?f=%5Cbf%20AH%3D%5Csqrt%7BHF%5E2%2BFA%5E2%7D%5Cimplies%20AH%3D%5Csqrt%7B9000%7D%20%5C%5C%5C%5C%5C%5C%20AB%3D%5Csqrt%7BAS%5E2%2BSB%5E2%7D%5Cimplies%20AB%3D%5Csqrt%7B7200%7D%20%5C%5C%5C%5C%5C%5C%20HB%20%3D%20AH%20%3D%20%5Csqrt%7B9000%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7Bperimeter%7D%7B%5Csqrt%7B9000%7D%2B%5Csqrt%7B7200%7D%2B%5Csqrt%7B7200%7D%7D%5Cqquad%20%5Capprox%20%5Cqquad%20%5Cstackrel%7Bperimeter%7D%7B264.57%7D)
now, we can plug those values in the Heron's Area Formula to get its area.
![\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{9000}\\ b=\sqrt{7200}\\ c=\sqrt{7200}\\ s\approx 132.29 \end{cases} \\\\\\ A=\sqrt{132.29(132.29-\sqrt{9000})(132.29-\sqrt{7200})(132.29-\sqrt{7200})} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill A\approx 3337.289~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BHeron%27s%20area%20formula%7D%20%5C%5C%5C%5C%20A%3D%5Csqrt%7Bs%28s-a%29%28s-b%29%28s-c%29%7D%5Cqquad%20%5Cbegin%7Bcases%7D%20s%3D%5Cfrac%7Ba%2Bb%2Bc%7D%7B2%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D%5Csqrt%7B9000%7D%5C%5C%20b%3D%5Csqrt%7B7200%7D%5C%5C%20c%3D%5Csqrt%7B7200%7D%5C%5C%20s%5Capprox%20132.29%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D%5Csqrt%7B132.29%28132.29-%5Csqrt%7B9000%7D%29%28132.29-%5Csqrt%7B7200%7D%29%28132.29-%5Csqrt%7B7200%7D%29%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20A%5Capprox%203337.289~%5Chfill)
First, find the number of shaded blocks.
shaded blocks = 8 × 3
shaded blocks = 24
There are 24 shaded blocks
Second, find the number of the blocks
all blocks = 8 × 10
all blocks = 80
There are 80 blocks in total.
Third, write a fraction defining the shaded blocks compare to the blocks in total
fraction = 24/80
simplify
fraction = 3/10
Fourth, change the fraction into percent
Percent means per hundred. Change the denominator to 100
fraction = 3/10
percent = (3 × 10) / (10 × 10)
percent = 30/100
percent = 30%
The percentage of the blocks shaded in the picture is 30%
Answer:
P(x< 18) = 0.986
Step-by-step explanation:
Step 1: find the z-score using the formula, z = (x - µ)/σ
Where,
x = randomly chosen values = 18
µ = mean = 7
σ = standard deviation = 5
Proportion of the population that is less than 18 = P(x < 18)
Plug in the values into z = (x - µ)/σ, to get z-score.
z = (18 - 7)/5
z = 11/5 = 2.2
Step 2: Find P(x< 18) = P (z<2.2) using z-table.
The probability that corresponds with z-score calculated is 0.986.
Therefore,
P(x< 18) = 0.986
The algebraic expression for the word phrase "the product of a number and 3" as a variable expression is 3z
<h3>How to write an algebraic expression for the word phrase "the product of a number and 3" as a variable expression?</h3>
The word phrase is given as:
"the product of a number and 3"
Represent the number with z
So, the word phrase can be rewritten as:
"the product of a number z and 3"
The product of a number z and 3 is represented as:
z * 3
When evaluated,, the expression becomes
z * 3 = 3z
Hence, the algebraic expression for the word phrase "the product of a number and 3" as a variable expression is 3z
Read more about algebraic expression at
brainly.com/question/4344214
#SPJ1
