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4 years ago

Evaluate the expression when x=3 and y= -2

Mathematics

1 answer:

Paul [167]4 years ago

5 0

Answer:

-11

Step-by-step explanation:

To evaluate an expression with variables, replace the variables with what the question tells you to.

Replace "x" with 3. Replace "y" with -2.

-x + 4y

= -(3) + 4(-2) Multiply 4 and -2 first to get -8

= (-3) + (-8) Add normally. -3 + (-8) is the same as (-3) - 8.

= -11 Answer

Therefore the solution is -11.

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Help! Include working out in answer please :)

Sliva [168]

N represents the number in the sequence.
1st term: (1-2)/1^2=-1/1=-1
2nd term: (2-2)/2^2=0/4=0
3rd term: (3-2)/3^2=1/9

3 0

3 years ago

HELP ME PLEASE!! ILL MARK U BRANLIEST!

ser-zykov [4K]

Answer:

Part A- B Part B- 50

Step-by-step explanation:

4x=200 if x represented 50 than 4X50 = 200 !

7 0

3 years ago

Read 2 more answers

1) Ten samples of a process measuring the number of returns per 100 receipts were taken for a local retail store.

densk [106]

Answer:

E) .0863

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

For this case we can find the sample proportion for each observation with the following formula:

$\hat p = \frac{X}{n}$

Where X represent the number of returns and n =100 for each case since the standard value used. Using the last formula we got:

$p_1 = \frac{10}{100}=0.1$

$p_2 = \frac{9}{100}=0.09$

$p_3 = \frac{11}{100}=0.11$

$p_4 = \frac{7}{100}=0.07$

$p_5 = \frac{3}{100}=0.03$

$p_6 = \frac{12}{100}=0.12$

$p_7 = \frac{8}{100}=0.08$

$p_8 = \frac{4}{100}=0.04$

$p_9 = \frac{6}{100}=0.06$

$p_{10} = \frac{11}{100}=0.11$

And now witht those values we can find the sample mean of proportions with the following formula:

$hat p = \frac{\sum_{i=1}^n p_i}{n}= \frac{0.1+0.09+0.11+0.07+0.03+0.12+0.08+0.04+0.06+0.11}{10}=0.081$

And we can find the standard error with the following formula:

$SE= \sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.081(1-0.081)}{10}}=0.0863$

So then the best option on this case is given by:

E) .0863

6 0

3 years ago

Po is an oncologist with seven patients in their care. the probability that a patient will survive five years after being diagno

alekssr [168]

Answer:

0.0923 = 9.23% probability that four of the patients are still alive after five years.

Step-by-step explanation:

For each patient, there are only two possible outcomes. Either they are still alive after five years, or they are not. The probability of a patient being alive is independent of any other patient, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Po is an oncologist with seven patients in their care.

This means that $n = 7$

The probability that a patient will survive five years after being diagnosed with stage three breast cancer is 0.82.

This means that $p = 0.82$

What is the probability that four of the patients are still alive after five years?

This is P(X = 4). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{7,4}.(0.82)^{4}.(0.18)^{3} = 0.0923$

0.0923 = 9.23% probability that four of the patients are still alive after five years.

5 0

3 years ago

(a) Find the value of xif 10, x and 30 are in A.P.

jekas [21]

Part A

<h3>Answer: x = 20</h3>

--------------------------------

Explanation:

AP = arithmetic progression, which is another way of saying arithmetic sequence.

Let d = common difference

first term = 10
second term = first+d = 10+d = x
third term = second+d = x+d = 30

Apply substitution like so

x+d = 30

10+d+d = 30 ... replaced x with 10+d

10+2d = 30

2d = 30-10

2d = 20

d = 20/2

d = 10 is the common difference

Therefore,

x = 10+d = 10+10 = 20

The AP {10,x,30} updates to {10,20,30}. We see the gap between terms is 10 units. All AP's have the same gap width between adjacent terms.

=============================================================

Part B

<h3>Answers: x = 10 and y = 14 </h3>

--------------------------------

Explanation:

We use the same ideas mentioned back in part A.

d = common difference = unknown for now

first term = 6
second term = first+d = 6+d = x
third term = second+d = x+d = (6+d)+d = 6+2d = y
fourth term = third+d = y+d = (6+2d)+d = 6+3d = 18

Hopefully you can see how each term builds up to form the next one. We'll solve that last equation like so

6+3d = 18

3d = 18-6

3d = 12

d = 12/3

d = 4

So we then can say:

x = 6+d = 6+4 = 10
y = 6+2d = 6+2(4) = 6+8 = 14

The arithmetic sequence {6,x,y,18} updates to {6,10,14,18}. There's a gap of 4 between each adjacent term.

6 0

3 years ago