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Svet_ta [14]
4 years ago
10

What value of x is in the solution set of 2(3x - 1) = 4x - 6? 0-10 0-5 -1

Mathematics
1 answer:
maxonik [38]4 years ago
3 0

Step-by-step explanation:

6x-2 = 4x-6

2x+4= 0

x= -2

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An athlete trains on a circular track of radius 40 m and jogs 10 laps each day, 5 days a week. How far does he jog each week? Ro
mariarad [96]
Length of a circle=2πr
r=40 m
Length of the circular track=2*π*40m=80π m.

The athelete jogs each day 10 laps
1 lap-------------------80π m
10 laps-------------   x
x=(10 laps * 80π m) / 1 lap=800π m

Calculate how far he jog eache week (5 days).
1 day------------------800π m
5 days----------------   x
x=(5 days*800π m)/1day=4000π m=4000m * 3.141592...≈12566,37 m

solution: 12566 m

5 0
3 years ago
Polygon ABCDE is the first in a pattern for a high school art project. The polygon is transformed so that the image of A’ is at
djyliett [7]

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4.21

Step-by-step explanation:

Because 2.1 is also equivalent to  -4.2 which also is equivalent to D'.

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3 years ago
Me ayudan por favor con estos ejercicios de álgebra con la 15, 16 y 17
SashulF [63]
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7 0
3 years ago
20 minus 3.05 plus (-9.5) =<br><br><br><br> HELP ME PLEASE
Mazyrski [523]

Answer:

7.45

Step-by-step explanation:

20 - 3.05 + (-9.5)

16.95 + (-9.5)

16.95 - 9.5

7.45

7 0
3 years ago
Read 2 more answers
Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno
Roman55 [17]

Answer:

a) P(X>np+3\sqrt{np(1-p)}=0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= \sqrt{0.198} ≈ 0.445

a) P(X>np+3\sqrt{np(1-p)}=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - \frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}-\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - \frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)} - \frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)=\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}=0.349

P(Y>1)=1-0.349=0.651

3 0
3 years ago
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