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Evgesh-ka [11]
3 years ago
14

Is every natural number a rational number.

Mathematics
1 answer:
kolezko [41]3 years ago
8 0
True
False
False
True
False
False
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For problems 1–3, identify the terms, coefficients, constants, and factors of the given expressions.
Volgvan

Answer:

1 4 ^2  +  2 - 9

7 0
3 years ago
Read 2 more answers
F(-4) for f (x) = -4x+9<br>f(-4)=?
Dovator [93]

F(x)= -4x+9

Multiply f*x

xF=9+4x

Solve for X: -4x+xF=9+4x-4x=0

Combine like terms: -4x+4x=0

-4x+xF=9+0= 9

-9+-4x +xF = 0.

3 0
3 years ago
16
Tanya [424]

Answer:

2.50

Step-by-step explanation:

Given

(x_1,y_1) = (-1,0.8)

(x_2,y_2) = (0,2)

(x_3,y_3) = (1,5)

(x_4,y_4) = (2,12.5)

Required

The rate of change (b)

The above graph is represented as:

y = ab^x

For: (x_2,y_2) = (0,2);

We have:

y = ab^x

2 = a * b^0

2 = a *1

2 = a

a = 2

For (x_3,y_3) = (1,5),

We have:

y = ab^x

5 = a * b^1

5 = a * b

Substitute a = 2

5 = 2 * b

Divide by 2

2.5 = b

b = 2.5

<em>Hence, the rate of change is 2.50</em>

7 0
3 years ago
A large electronic office product contains 2000 electronic components. Assume that the probability that each component operates
KIM [24]

Answer:

The probability is 0.971032

Step-by-step explanation:

The variable that says the number of components that fail during the useful life of the product follows a binomial distribution.

The Binomial distribution apply when we have n identical and independent events with a probability p of success and a probability 1-p of not success. Then, the probability that x of the n events are success is given by:

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

In this case, we have 2000 electronics components with a probability 0.005 of fail during the useful life of the product and a probability 0.995 that each component operates without failure during the useful life of the product. Then, the probability that x components of the 2000 fail is:

P(x)=\frac{2000!}{x!(2000-x)!}*0.005^{x}*(0.995)^{2000-x}     (eq. 1)

So, the probability that 5 or more of the original 2000 components fail during the useful life of the product is:

P(x ≥ 5) = P(5) + P(6) + ... + P(1999) + P(2000)

We can also calculated that as:

P(x ≥ 5) = 1 - P(x ≤ 4)

Where P(x ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4)

Then, if we calculate every probability using eq. 1, we get:

P(x ≤ 4) = 0.000044 + 0.000445 + 0.002235 + 0.007479 + 0.018765

P(x ≤ 4) = 0.028968

Finally, P(x ≥ 5) is:

P(x ≥ 5) = 1 - 0.028968

P(x ≥ 5) = 0.971032

3 0
3 years ago
Develop a 95% confidence interval for the expected value of y when x = 8. estimate the standard deviation of an individual value
zloy xaker [14]

I don’t know what the answer is I wish I could help

3 0
3 years ago
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