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BartSMP [9]
4 years ago
14

What is the value of three in 34,525

Mathematics
1 answer:
LUCKY_DIMON [66]4 years ago
7 0
5=ones
2=tens
5=hundreds
4=thousands
3=ten thousands
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What is 3(x+4) I need the answer ASAP!!!
Luda [366]

Answer:

3x+12

Step-by-step explanation:

3(x+4)

Use the distributive property to multiply 3 by x+4.

3x+12

4 0
3 years ago
Which line has a y-intercept of –2?a)
Elden [556K]

Answer:

P

Step-by-step explanation:

It passes through -2 on the y-axis

HOPE THIS HELPS! (:

8 0
3 years ago
I need help with this question asap!!
Sholpan [36]

Answer:

1/4= 25%

Step-by-step explanation:

6 0
3 years ago
A box contains four 75 W lightbulbs, three 60 W lightbulbs, and three burned-out lightbulbs. Two bulbs are selected at random fr
Law Incorporation [45]

Answer:

a. 0.689

b. 0.8

c. 0.427

Step-by-step explanation:

The given scenario indicates hyper-geometric experiment because because successive trials are dependent and probability of success changes on each trial.

The probability mass function for hyper-geometric distribution is

P(X=x)=kCx(N-k)C(n-x)/NCn

where N=4+3+3=10

n=2

k=4

a.

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

The probability mass function for hyper-geometric distribution is

P(X=x)=kCx(N-k)C(n-x)/NCn

P(X=0)=4C0(6C2)/10C2=15/45=0.311

P(X>0)=1-P(X=0)=1-0.311=0.689

P(X>0)=0.689

b.

The mean of hyper-geometric distribution is

μx=nk/N

μx=2*4/10=8/10=0.8

c.

The variance of hyper-geometric distribution is

σx²=nk(N-k).(N-n)/N²(N-1)

σx²=2*4(10-4).(10-2)/10²*9

σx²=8*6*8/900=384/900=0.427

6 0
3 years ago
Find the vertical and horizontal asymptote if they exist
tresset_1 [31]

Hello!

Vertical asymptotes are determined by setting the denominator of a rational function to zero and then by solving for x.

Horizontal asymptotes are determined by:

1. If the degree of the numerator < degree of denominator, then the line, y = 0 is the horizontal asymptote.

2. If the degree of the numerator = degree of denominator, then y = leading coefficient of numerator / leading coefficient of denominator is the horizontal asymptote.

3. If degree of numerator > degree of denominator, then there is an oblique asymptote, but no horizontal asymptote.

To find the vertical asymptote:

2x² - 10 = 0

2(x² - 5) = 0

(x - √5)(x + √5) = 0

x = √5 and x = -√5

Graphing the equation, we realize that x = -√5 is not a vertical asymptote, so therefore, the only vertical asymptote is x = √5.

To find the horizontal asymptote:

If the degree of the numerator < degree of denominator, then the line, y = 0 is the horizontal asymptote.

Therefore, the horizontal asymptote of this function is y = 0.

Short answer: Vertical asymptote: x = √5 and horizontal asymptote: y = 0

7 0
3 years ago
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