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storchak [24]
2 years ago
10

2x+y=25. 3y=2x-13 solve by elimination

Mathematics
1 answer:
Mila [183]2 years ago
3 0
What  are you trying to find X or Y ?

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Four different prime numbers, each less than 20, are multiplied together. What is greatest possible result?
valentinak56 [21]

Answer:

46,189

Step-by-step explanation:

The prime numbers that are less than 20 are :

1,2,3,5,7,11,13,17,19

to get the greatest value, we multiply the four numbers with the largest values i.e

11 x 13 x 17 x 19 = 46,189

8 0
2 years ago
math help ! Will give branliest !! At a car and truck dealership, the probability that a vehicle is white is 0.25 . The probabil
defon

Answer:

As per the statement:

Let  Event A represents the probability that a vehicle is white and Event B  represents the probability that it is a pick up truck .

then;

P(A) = 0.25 and P(B) = 0.15

It is also given that:

The Probability that it is a white pick up truck is 0.06.

⇒P(A \cap B) = 0.06 where A \cap B represents that it is white pick up truck.

We have to find the the probability that the vehicle is white {given that the vehicle is a pickup truck}

We use the formula:

P(A/B) = \frac{P(A \cap B)}{P(B)}

where

A/B represents that the pick up truck vehicle is white.

Substitute the given values we get;

P(A/B) = \frac{0.06}{0.15} = 0.4

Therefore,  the probability that the vehicle is white, given that the vehicle is a pickup truck is 0.4

7 0
2 years ago
Which situation involves descriptive statistics?
zmey [24]
They all involve descriptive statistics.
8 0
2 years ago
Read 2 more answers
Announcements for 84 upcoming engineering conferences were randomly picked from a stack of IEEE Spectrum magazines. The mean len
Marina86 [1]

Answer:

Step-by-step explanation:

Hello!

You need to construct a 95% CI for the population mean of the length of engineering conferences.

The variable has a normal distribution.

The information given is:

n= 84

x[bar]= 3.94

δ= 1.28

The formula for the Confidence interval is:

x[bar]±Z_{1-\alpha/2}*(δ/n)

Lower bound(Lb): 3.698

Upper bound(Ub): 4.182

Error bound: (Ub - Lb)/2 = (4.182-3.698)/2 = 0.242

I hope it helps!

6 0
2 years ago
The thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters. Determine the foll
Mrrafil [7]

Answer

given,

thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters.

X = U[0.95,1.05]           0.95≤ x ≤ 1.05

the cumulative distribution function of flange

F(x) = P{X≤ x}=\dfrac{x-0.95}{1.05-0.95}

                     =\dfrac{x-0.95}{0.1}

b) P(X>1.02)= 1 - P(X≤1.02)

                   = 1- \dfrac{1.02-0.95}{0.1}

                   = 0.3

c) The thickness greater than 0.96 exceeded by 90% of the flanges.

d) mean = \dfrac{0.95+1.05}{2}

              = 1

   variance = \dfrac{(1.05-0.95)^2}{12}

                  = 0.000833

4 0
2 years ago
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