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DiKsa [7]
2 years ago
7

When Ricardo was years old, he was 56

Mathematics
1 answer:
Anestetic [448]2 years ago
8 0
6% should be the answer
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QUESTION 7<br> Solve for x.<br> (3x + 32)<br> (5x-8)<br> O a. 3<br> b. 12<br> O c. 20<br> O d. 19.5
Alchen [17]

Answer:

The answer is x = 20

Step-by-step explanation:

If you set both equations equal to each other and solve for x, you will get 20.

3x+32 = 5x-8        add 8 to both sides

3x+40 = 5x           subtract 3x from both sides

40 = 2x                 divide by 2 on both sides

20 = x

7 0
3 years ago
The total cost of 4 starfish was $7.20. Write and solve an equation to find how much one starfish cost.
Lyrx [107]

Answer:

$1.80

Step-by-step explanation:

4x = 7.20

x = 1.80

7 0
2 years ago
43/21 decimal rounded to the nearest hundredth
Step2247 [10]

Answer:

2.05

Step-by-step explanation:

4 0
2 years ago
There are several scenarios described below. For each of them, do the following (note: R.V. means random variable) (1) Define th
frutty [35]

Answer:

a) The Ohio Bureau of Motor Vehicles states that 7 out of 8 people pass the written driver’s test.

Let X be the number of test given by the test taker to pass out.

So X~Geometric(p) where p=Probability that for a particular test anyone will pass the written test=7/8

and here support of X be equal to 1,2,3,..... i.e X is a Natural number

So ,probability that he will pass the written test in fewer than 4 tries

=P(X<4)

=\sum _{x=0}^{3}p(1-p)^{x-1}

b) LAIMO manufacturing company makes parts for the auto industry. Approximately 3% of the parts it makes are defective.

So let X=number of non defective parts sampled before the 3rd defective part is sampled

then X~Negative Binomial(r,p) where here r=3 and p=Probability that a randomly selected part is defective= 0.03

where support of X is {0,1,2,3,...}

So the probability that the third defective part is the 20th one sampled.

P(X=20-3=17)

=\binom{r+16}{17}p^r(1-p)^{16}

c) A BigMart store is going to hire 3 new cashiers. It has 18 applicants (10 male, 8 female) for these 3 cashier jobs.

So let X be number of female cashier appointed.

Here X~Hypergeometric(3,8,18) where

f(x)=P(X=x)

=\left\{\begin{matrix} \frac{\binom{8}{x}.\binom{10}{3-x}}{\binom{18}{3}} & ,x=0,1,2,3\\ 0 & ,otherwise \end{matrix}\right.

So the probability that none of the positions are filled by females

=P(X=0)

d) A gardener is inspecting the fall flowers in her garden. She notices, on average, 4 bugs on a flower. She randomly picks one flower from her garden.

Let X be the numbers of bugs on that flower

So X follows Poisson distribution with mean 4 where support of X is {0,1,2,3.....}

So the probability that the flower she picked has at least one bug on it

=P(X\geq 1)=1-P(X=0)

=1-e^{-4}\frac{4^x}{x!}|_{x=0}

e) A student is taking a true/false test that consists of 15 questions. Based on past performance the student has approximately a 70% chance of getting any individual question correct.

So let X be the number of questions that are correct among those 15 questions.

so X~Binomial(n,p) where n=15 and p=Probability that he get an individual question correct =0.7

where support of X be {0,1,2,3,...,15}

So the probability that the student gets at least 60% of the questions on the test correct or 15x60%=9 questions are correct

=P(X\geq9)

=\sum _{x=9}^{15}\binom{15}{x}p^x(1-p)^{15-x}

f) A certain radio station’s phone lines are busy approximately 95% of the time when trying to call during a contest.

Let X denotes the number of calls to get into the contest.

So X~Geometric(p) where p=Probability that in a call I get through into the contest=1-0.95=0.05

support of x={1,2,3,....}

So the probability that the 4 th time you call is the 1st time you get through during a contest.

=P(X=4)

=p(1-p)^4

4 0
2 years ago
I need help!!! Idk what the answer is
yanalaym [24]

Translations, reflections, and rotations preserve congruency. Dilations do not?

3 0
3 years ago
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