All categories

3 years ago

Which is greatest 9/1 > 8/10

Mathematics

1 answer:

Andrew [12]3 years ago

5 0

Answer:

9/1 is the greatest

9/1 = 9

8/10 = 0.8

9/1 > 8/10

You might be interested in

Does anyone know what this answer this

serg [7]

I believe the answer is 7 since 8x squared is 64x and subtract 4x=60x
60x/16x multiplied by 5/x-5. but i am not sure since i do not know the value of X sorry buddy, just trying to help

4 0

3 years ago

Subtract 1/4w + 4 from 1/2w + 3

Aleks04 [339]

I think it should be w+28/4 if thats one of the answer choices.

4 0

3 years ago

Mia wants to buy 3 notebooks for $1.29 each. What is the expression for the total cost?

ololo11 [35]

Step-by-step explanation:

Mia want to buy 3 notebook for $1.29 each ,

then total cost =3×1.29

hence, 3×29

1\6+1/12+2/6=2+1+4/12

=7/12

8 0

3 years ago

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection = </em>

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

5 0

3 years ago

ABCD is a parallelogram. Find the values of x and y.

Komok [63]

Answer:

x = 8; y = 29

Step-by-step explanation:

In a parallelogram, opposite angles are congruent.

8x = 7x + 8

x = 8

4y = y + 87

3y = 87

y = 29

6 0

3 years ago

Read 2 more answers