All categories

2 years ago

If mc020-1.jpg and mc020-2.jpg, what is the domain of mc020-3.jpg?

Mathematics

2 answers:

svp [43]2 years ago

5 0

Answer:

we have

f(x)=x+7

g(x)=\frac{1}{x-13}

(fog)(x)=[\frac{1}{x-13}]+7

we know that

the denominator cannot be zero, therefore the value of x cannot be 13

The domain are all real numbers except the number 13

therefore

the answer is the option

{x|≠13}

ludmilkaskok [199]2 years ago

3 0

we have

$f(x)=x+7$

$g(x)=\frac{1}{x-13}$

$(fog)(x)=[\frac{1}{x-13}]+7$

we know that

the denominator cannot be zero, therefore the value of x cannot be $13$

The domain are all real numbers except the number $13$

therefore

<u>the answer is the option</u>

{x|≠13}

You might be interested in

A student repeatedly measures the mass of an object using a mechanical balance and gets the following values: 560 g, 562 g, 556

MrRissso [65]

Answer: 2.76 g

Step-by-step explanation:

The formula to find the standard deviation:-

$\sigma=\sqrt{\dfrac{\sum(x_i-\overline{x})^2}{n}}$

The given data values : 560 g, 562 g, 556 g, 558 g, 560 g, 556 g, 559 g, 561 g, 565 g, 563 g.

Then, $\overline{x}=\dfrac{\sum_{i=1}^{10} x_i}{n}\\\\\Rightarrow\ \overline{x}=\dfrac{560+562+556+558+560+556+559+561+565+563}{10}\\\\\Rightarrow\ \overline{x}=\dfrac{5600}{10}=560$

Now, $\sum_{i=1}^{10}(x_i-\overline{x})^2=0^2+2^2+(-4)^2+(-2)^2+0^2+(-4)^2+(-1)^2+1^2+5^2+3^2\\\\\Rightarrow\ \sum_{i=1}^{10}(x_i-\overline{x})^2=76$

Then, $\sigma=\sqrt{\dfrac{76}{10}}=\sqrt{7.6}=2.76$

Hence, the standard deviation of his measurements = 2.76 g

6 0

2 years ago

A doctor observes a graph that shows the electrical activity (in volts) of the heart of a patient over a period of time (in seco

hodyreva [135]

Answer:

90 beats per minute

Step-by-step explanation:

By looking at the graph, we see that exactly at second 6, the 9th beat occurs. We can take that to beats/min by multiplying that relation by 10 (since there are ten 6sec in one minute).

9beats every 6sec * 10 = 90beats/min

6 0

2 years ago

Read 2 more answers

In circle N, KL ≅ ML. Circle N is shown. Line segments N J, N M, N L, and N K are radii. Lines are drawn to connect each point o

Sergeu [11.5K]

Answer:

$arc\ JM=132^o$

Step-by-step explanation:

The picture of the question in the attached figure

step 1

we know that

$arc\ JM+arc\ ML+arc\ LK+arc\ KJ=360^o$ ----> by complete circle

substitute the given values

$(13x+2)^o+(8x-3)^o+(7x+7)^o+(5x+24)^o=360^o$

solve for x

$(33x+30)^o=360^o\\33x=360-30\\33x=330\\x=10$

step 2

Find the measure of arc JM

$arc\ JM=(13x+2)^o$

substitute the value of x

$arc\ JM=(13(10)+2)^o=132^o$

8 0

2 years ago

Read 2 more answers

If 300 boxes of medical sponges cost $25.00, how many boxes can you purchase for $0.50?

nikdorinn [45]

You can purchase 6 boxes of medical sponges for $0.50.

Step 1:
Divide 300 by 25.00
300/25= 12
So you get 12 boxes for 1 dollar.
We have 50 cents.

Step 2:
Multiply 12 by 0.50
12*.5 = 6
So you can buy 6 boxes for $0.50.

Hope this helps you! (:
-Hamilton1757

4 0

1 year ago

Read 2 more answers

Let Y denote a geometric random variable with probability of success p. a Show that for a positive integer a, P(Y > a) = qa .

lakkis [162]

Answer:

a) For this case we can find the cumulative distribution function first:

$F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k$

So then by the complement rule we have this:

$P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a$

b) $P(Y>a)= q^a$

$P(Y>b) = q^b$

So then we have this using independence:

$P(Y> a+b) = q^{a+b}$

We want to find the following probability:

$P(Y> a+b |Y>a)$

Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

And we see that if a = 2 and b=5 we have:

$P(Y> 2+5 | Y>2) = P(Y>5)$

c) For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

If we define the random of variable Y we know that:

$Y\sim Geo (1-p)$

Part a

For this case we can find the cumulative distribution function first:

$F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k$

So then by the complement rule we have this:

$P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a$

Part b

For this case we can use the result from part a to conclude that:

$P(Y>a)= q^a$

$P(Y>b) = q^b$

So then we have this assuming independence:

$P(Y> a+b) = q^{a+b}$

We want to find the following probability:

$P(Y> a+b |Y>a)$

Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

And we see that if a = 2 and b=5 we have:

$P(Y> 2+5 | Y>2) = P(Y>5)$

Part c

For this case we use independent identical and with the same distribution experiments.

8 0

2 years ago