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3 years ago

What is the slope, m, and the y-intercept of the line that is graphed below?

Mathematics

1 answer:

Pepsi [2]3 years ago

6 0

Answer:

Slope: 1

Y-intercept: (0,3)

Step-by-step explanation:

The y intercept is when the slope reaches the y-axis line. In this case, it is given to us. Anything that is formed like this: (0, y) is the y-intercept.

Y intercept: (0, 3)

For slope, you can use the formula rise over run. $\frac{Rise}{Run}$

From the picture, I have drawn the rise over run, which is $\frac{3}{3}$ , which is also 1.

Slope: 1

Hope this helped.

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A series of 384 consecutive odd integers has a sum that is a perfect fourth power of a positive

777dan777 [17]

Using an arithmetic sequence, it is found that the smallest possible sum for the series is of 20 736, given by option B.

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In an arithmetic sequence, the difference of consecutive terms is always the same, called common difference.
The nth term of an arithmetic sequence is given by:

$a_n = a_1 + (n-1)d$

The sum of the first n terms is given by:

$S_n = \frac{n(a_1+a_n)}{2}$

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

The set of odd integers is an arithmetic sequence with common difference 2, thus $d = 2$ .
384 terms, thus $n = 384$
The last term is: $a_{384} = a_1 + 383(2) = a_1 + 766$

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

The sum of the 384 terms is:

$S_{384} = \frac{384(a_1 + a_1 + 766)}{2} = 192(2a_1 + 766) = 384a_1 + 147072$

Now, for each option, we have to test if it generates an odd $a_1$ .

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

Option d: Sum of 1296, thus, $S_{384} = 1296$ , solve for $a_1$

$384a_1 + 147072 = 1296$

$a_1 = \frac{1296 - 147072}{384}$

$a_1 = -379.6$

Not an integer, so not the answer.

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

Option c: Test for 10000.

$384a_1 + 147072 = 10000$

$a_1 = \frac{10000 - 147072}{384}$

$a_1 = -356.9$

Not an integer, so not the answer.

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

Option b: Test for 20736.

$384a_1 + 147072 = 20736$

$a_1 = \frac{20736 - 147072}{384}$

$a_1 = -329$

Integer an odd, thus, option b is the answer.

A similar problem is given at brainly.com/question/16720434

8 0

2 years ago

Raphael graphed the functions g(x) = x + 2 and f(x) = x – 1. How many units below the y-intercept of g(x) is the y-intercept of

muminat

These functions are in the form y=mx+b
Therefore, we know that in g(x) = x+2, the y-intercept is 2.
For f(x) = x-1, the y-intercept is -1.
Now find the difference between them.
2-(-1)= 3
The y-intercept for f(x) is 3 units below g(x)

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2 years ago

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What is the product?<br> (4%)(-3x0)(-7x)

Degger [83]

The answer to your question is -28

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2 years ago

Line k is parallel to line l

pickupchik [31]

Angle 1. Because angle 4 and 1 are vertical angles and vertical angles are congruent.

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Among persons donating blood to a clinic, 85% have Rh+ blood (that is, the Rhesus factor is present in their blood.) Six people

Leona [35]

Answer:

a) There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) There is a 22.36% probability that at most four of the six have Rh+ blood.

c) There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

Step-by-step explanation:

For each person donating blood, there are only two possible outcomes. Either they have Rh+ blood, or they do not. This means that we use the binomial probability distribution to solve this problem.

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 combinatios 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.

In this problem we have that:

$p = 0.85, n = 6$ .

a) fine the probability that at least one of the five does not have the Rh factor.

Either all six have the factor, or at least one of them do not. The sum of the probabilities of these events is decimal 1. So:

$P(X < 6) + P(X = 6) = 1$

$P(X < 6) = 1 - P(X = 6)$

In which:

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

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

$P(X < 6) = 1 - P(X = 6) = 1 - 0.3771 = 0.6229$

There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) find the probability that at most four of the six have Rh+ blood.

Either more than four have Rh+ blood, or at most four have. So

$P(X \leq 4) + P(X > 4) = 1$

$P(X \leq 4) = 1 - P(X > 4)$

In which

$P(X > 4) = P(X = 5) + P(X = 6)$

$P(X = 5) = C_{6,5}.(0.85)^{5}.(0.15)^{1} = 0.3993$

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

$P(X > 4) = P(X = 5) + P(X = 6) = 0.3993 + 0.3771 = 0.7764$

$P(X \leq 4) = 1 - P(X > 4) = 1 - 0.7764 = 0.2236$

There is a 22.36% probability that at most four of the six have Rh+ blood.

c) The clinic needs six Rh+ donors on a certain day. How many people must donate blood to have the probability of obtaining blood from at least six Rh+ donors over 0.95?

With 6 donors:

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

37.71% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 7 donors:

$P(X = 6) = C_{7,6}.(0.85)^{6}.(0.15)^{1} = 0.3960$

0.3771 + 0.3960 = 0.7764 = 77.64% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 8 donors

$P(X = 6) = C_{8,6}.(0.85)^{6}.(0.15)^{2} = 0.2376$

0.3771 + 0.3960 + 0.2376 = 1.01 = 101% probability of obtaining blood from at least six Rh+ donors over 0.95.

There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

5 0

3 years ago