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3 years ago

PLEASE HELP. Write in algebraic form: “the square root of the sum of a and b”

Mathematics

2 answers:

Juli2301 [7.4K]3 years ago

7 0

Answer:

The answer is (a + b )^2

Step-by-step explanation:

chubhunter [2.5K]3 years ago

5 0

Write in algebraic form: “the square root of the sum of a and b”

answer: $\sqrt{a+b}$

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A picture is mounted on a frame measuring 25 centimeters by

makvit [3.9K]

Answer:

I think it's 1,125 and 4485

Step-by-step explanation:

25x13x3

115x13x3

3 0

3 years ago

If y varies directly with x and y = 25 when x = 5, what is the value of k? how does this work when y = kx?

Savatey [412]

If y = 25 and x = 5

25 = 5k

k = 5

3 0

3 years ago

Read 2 more answers

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

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

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.

This means that,

Probability that at least one of them needs replacement at the first inspection =

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

5 0

3 years ago

35=7x. please help me fasttt

AlladinOne [14]

Answer:

x=5

Step-by-step explanation:

35 divided by 7 is 5, so in that case, 7 times 5 is 35.

3 0

3 years ago

Use the limit theorem and the properties of limits to find the limit. Picture provided below

sashaice [31]

Answer:

b. 1/2

Step-by-step explanation:

lim (x -3)(x +2)

x-->-∞ ---------------

2x^2 + x +1

= lim (x^2 -3x +2x - 6)

x-->-∞ -----------------------

2x^2 + x +1

= lim (x^2 -x - 6)

x-->-∞ -----------------------

2x^2 + x +1

When we plug in x = -∞, we get indeterminate form.

Now we have to use the L'hospital rule.

d/dx (x^2 - x - 6) = 2x -1

d/dx (2x^2 + x + 1) = 4x + 1

Now apply the limit

lim (2x - 1) / (4x + 1)

x--->-∞

Here we have to degree of the numerator and the denominator of the same. In this case, if x --> -∞, we get the result as the coefficient of the leading term as the result.

According to the rule, we get

= 2/4

Which can simplified as 1/2

The answer is 1/2

Hope this will helpful.

Thank you.

3 0

3 years ago

Read 2 more answers