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Ierofanga [76]
3 years ago
15

18+3x=-10+x what is x

Mathematics
2 answers:
azamat3 years ago
8 0
18+ 3x= -10 + x
18 + 2x = -10
2x= -28
x = -14
son4ous [18]3 years ago
7 0
18+3x=-10+x
3x-x=-10-18
2x=-28
x=-14
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If PQR is a triangle,given that /PQ\=15m, /QR\=17m and /RP\=18m.Find the area of the triangle to the nearest whole number.
nydimaria [60]

Answer:

118.3 m²

Step-by-step explanation:

Step 1. Calculate the <em>semiperimeter</em> (s).

s = (p + q + r)/2

s = (17 + 18+ 15)/2  

s = 50/2

s = 25 m

===============

Step 2. Calculate the <em>area</em> (A).

Use <em>Heron’s formula</em>:

A = \sqrt{s(s-p)(s-q)(s-r)}

A = \sqrt{25(25-17)(25-18)(25-15)}

A = \sqrt{25\times8\times7\times10}

A = \sqrt{14 000}

A = 20\sqrt{35}

A = 118.3 m²

3 0
3 years ago
Choose Yes or No to tell whether the Addition Property of Inequality can be used to solve each statement.
balu736 [363]

Answer:

yes,no, no, yes

Step-by-step explanation:

6 0
3 years ago
Sarah is doing research on the average age of first-year resident physicians. She wants her estimate to be accurate to within 1
WINSTONCH [101]

Answer: The number of first-year residents she must survey to be 95% confident= 263

Step-by-step explanation:

When population standard deviation (\sigma)  is known and margin of error(E) is given, then the minimum sample size (n) is given by :-

n=(\dfrac{z^*\sigma}{E})^2, z* = Two-tailed critical value for the given confidence interval.

For 95% confidence level , z* = 1.96

As, \sigma = 8.265, E = 1

So, n= (\dfrac{1.96\times8.265}{1})^2 =(16.1994)^2\\\\= 262.42056036\approx263\ \ \ [\text{Rounded to the next integer}]

Hence, the number of first-year residents she must survey to be 95% confident= 263

7 0
3 years ago
The prior probabilities for events A1 and A2 are P(A1) = 0.35 and P(A2) = 0.50. It is also known that P(A1 ∩ A2) = 0. Suppose P(
forsale [732]

Answer:

Step-by-step explanation:

Hello!

Given the probabilities:

P(A₁)= 0.35

P(A₂)= 0.50

P(A₁∩A₂)= 0

P(BIA₁)= 0.20

P(BIA₂)= 0.05

a)

Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)

Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.

b)

Considering that P(BIA)= \frac{P(AnB)}{P(A)} you can clear the intersection from the formula P(AnB)= P(B/A)*P(A) and apply it for the given events:

P(A_1nB)= P(B/A_1) * P(A_1)= 0.20*0.35= 0.07

P(A_2nB)= P(B/A_2)*P(A_2)= 0.05*0.50= 0.025

c)

The probability of "B" is marginal, to calculate it you have to add all intersections where it occurs:

P(B)= (A₁∩B) + P(A₂∩B)=  0.07 + 0.025= 0.095

d)

The Bayes' theorem states that:

P(Ai/B)= \frac{P(B/Ai)*P(A)}{P(B)}

Then:

P(A_1/B)= \frac{P(B/A_1)*P(A_1)}{P(B)}= \frac{0.20*0.35}{0.095}= 0.737 = 0.74

P(A_2/B)= \frac{P(B/A_2)*P(A_2)}{P(B)} = \frac{0.05*0.50}{0.095} = 0.26

I hope it helps!

5 0
3 years ago
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