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Dmitry [639]
3 years ago
7

Multiply (−4) × (−2) × (−5). A. −40 B. −8 C. 8 D. 40

Mathematics
2 answers:
slavikrds [6]3 years ago
8 0
The answer is A. hope it helps
OlgaM077 [116]3 years ago
6 0
A) -40
Hope this helps.
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The lifetime of a leaf blower is exponentially distributed with a mean of 5 years. If you buy a 12 year old leaf blower, what is
snow_tiger [21]

Answer:

1.967 × 10⁻¹¹

Step-by-step explanation:

This is a conditional probability problem

Probability that a 12 year old leaf blower works for 5 more years = Probabilty that a leaf blower works for 17 years, given that it has worked for 12 years = P(17|12)

P(17|12) = P(17 n 12)/P(12)

P(17 n 12) = P(17)

Exponential random variable is given as

f(x) = λ e^(-λ.x)

λ = rate parameter = 5 years per blower

x = variable whose probability is required = 17

f(17) = 17 e^(-5×17)

f(17) = 17 e⁻⁸⁵ = 2.067 × 10⁻³⁶

f(12) = 12 e^(-5×12)

f(12) = 12 e⁻⁶⁰ = 1.051 × 10⁻²⁵

P(17|12) = (2.067 × 10⁻³⁶)/(1.051 × 10⁻²⁵)

P(17|12) = 1.967 × 10⁻¹¹

8 0
3 years ago
13 is 65% of what number?​
umka21 [38]

Answer:

20

Step-by-step explanation:

set up equation

13/x=65/100

cross multiple: 1300=65x

divide 1300/65=20

5 0
3 years ago
PLEASE HELP<br><br> What is the value of x?
Naddika [18.5K]

Answer:

(3x-3)+[6(x-10)]

3x-3+6x-60

9x-63

/9 /9

x=7

5 0
3 years ago
Read 2 more answers
Y+x=88
jolli1 [7]

Answer:

X = 30

Step-by-step explanation:

1) Y + X = 88

2) 58 + X = 88

  -58          -58

------------------------

          X = 30

7 0
3 years ago
Find the value of kk for which the constant function x(t)=kx(t)=k is a solution of the differential equation 5t3dxdt+2x−2=05t3dx
uranmaximum [27]
Given the differential equation

5t^3 \frac{dx}{dt} +2x-2=0

The solution is as follows:

5t^3 \frac{dx}{dt} +2x-2=0 \\  \\ \Rightarrow5t^3 \frac{dx}{dt} =2-2x \\  \\ \Rightarrow \frac{5}{2-2x} dx= \frac{1}{t^3} dt \\  \\ \Rightarrow \int {\frac{5}{2-2x} } \, dx = \int {\frac{1}{t^3}} \, dt \\  \\ \Rightarrow- \frac{5}{2} \ln(2-2x)=- \frac{1}{2t^2} +A \\  \\ \Rightarrow\ln(2-2x)= \frac{1}{5t^2} +B\\  \\ \Rightarrow2-2x=Ce^{\frac{1}{5t^2}} \\  \\ \Rightarrow 2x=Ce^{\frac{1}{5t^2}}+2 \\  \\ \Rightarrow x=De^{\frac{1}{5t^2}}+1
3 0
3 years ago
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