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Step2247 [10]
2 years ago
9

Use the drawing tools to form the correct answer on the provided graph

Mathematics
1 answer:
Anna11 [10]2 years ago
4 0
Line graph is the only way i know to even come close to getting tha right
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Proof that the identities are equal to eachother​
Vikentia [17]

Answer:

Hello, hope this will help:)

6 0
3 years ago
Check image giving brainliest to first correct answer
Snowcat [4.5K]

Answer:

<em>7</em>

Step-by-step explanation:

<em>S</em><em>o</em><em>:</em><em>a</em><em>=</em><em>3</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em>b</em><em>=</em><em>(</em><em>-</em><em>2</em><em>)</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>c</em><em>=</em><em>2</em>

<em>S</em><em>u</em><em>b</em><em>s</em><em>t</em><em>i</em><em>t</em><em>u</em><em>t</em><em>e</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>n</em><em>u</em><em>m</em><em>b</em><em>e</em><em>r</em><em>s</em><em> </em><em>i</em><em>n</em><em>t</em><em>o</em><em> </em><em>e</em><em>a</em><em>c</em><em>h</em><em> </em><em>letters</em>

<em> </em><em>s</em><em>o</em><em> </em><em>a</em><em> </em><em>i</em><em>s</em><em> </em><em>r</em><em>e</em><em>p</em><em>r</em><em>e</em><em>s</em><em>e</em><em>n</em><em>t</em><em>e</em><em>d</em><em> </em><em>b</em><em>y</em><em> </em><em>3</em><em>,</em><em>c</em><em> </em><em>b</em><em>y</em><em> </em><em>2</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>b</em><em> </em><em>-</em><em>2</em>

<em> </em><em>t</em><em>h</em><em>e</em><em>r</em><em>e</em><em>f</em><em>o</em><em>r</em><em>e</em><em> </em><em>g</em><em>i</em><em>v</em><em>i</em><em>n</em><em>g</em><em> </em><em>u</em><em>s</em><em>:</em><em> </em><em> </em><em> </em><em> </em><em>2</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>3</em><em>+</em><em>(</em><em>-</em><em>2</em><em>)</em><em>=</em><em>9</em><em>+</em><em>(</em><em>-</em><em>2</em><em>)</em><em>=</em><em>7</em>

7 0
2 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
How to do number 15?
tatiyna
I think it would be 120 because it is 2 times 60 and they add up to 180 I may be wrong buts that’s what I would do
3 0
3 years ago
Michael had $550 in his savings account. He took out $30.75 every month for one year. What is the net change in Michael's accoun
kozerog [31]
$30.75 x 12 = 369
$550-$369 =$181
7 0
3 years ago
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