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blsea [12.9K]
3 years ago
11

Giving out brainliest! Please help me I’m struggling.

Mathematics
2 answers:
Naya [18.7K]3 years ago
7 0

Answer:

Step-by-step explanation:

quester [9]3 years ago
4 0

Answer:

3 is the slope of the line

Step-by-step explanation:

2 is the y-intercept:)

Hope I helped a little! I wish you lcuk with all my heart!

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⦁ Write an equation of a line in slope intercept form (y = mx + b) with slope 1/3 going through the point (-6, 2). Show your wor
Zepler [3.9K]

(\stackrel{x_1}{-6}~,~\stackrel{y_1}{2})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{\cfrac{1}{3}}(x-\stackrel{x_1}{(-6)}) \\\\\\ y-2=\cfrac{1}{3}(x+6)\implies y-2=\cfrac{1}{3}x+2\implies y=\cfrac{1}{3}x+4

3 0
2 years ago
Suppose that the population mean for income is $50,000, while the population standard deviation is 25,000. If we select a random
Fudgin [204]

Answer:

Probability that the sample will have a mean that is greater than $52,000 is 0.0057.

Step-by-step explanation:

We are given that the population mean for income is $50,000, while the population standard deviation is 25,000.

We select a random sample of 1,000 people.

<em>Let </em>\bar X<em> = sample mean</em>

The z-score probability distribution for sample mean is given by;

               Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean = $50,000

            \sigma = population standard deviation = $25,000

            n = sample of people = 1,000

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the sample will have a mean that is greater than $52,000 is given by = P(\bar X > $52,000)

  P(\bar X > $52,000) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{52,000-50,000}{\frac{25,000}{\sqrt{1,000} } } ) = P(Z > 2.53) = 1 - P(Z \leq 2.53)

                                                                    = 1 - 0.9943 = 0.0057

<em>Now, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.53 in the z table which has an area of 0.9943.</em>

Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.

5 0
3 years ago
Help please help me
BARSIC [14]

Step-by-step explanation:

if you're looking for x

x=180-42-51= 87°

and if you're looking for D

D= 180-42= 138°

7 0
3 years ago
Which equation represents a line that passes through (4,1/3) and has a slope of 3/4?
scoray [572]

Step-by-step explanation:

With this kind of problem, we're looking at an equation in the form

y - y1 = m(x - x1)

(m = slope)

so we can substitute m, y1, and x1 with the values we're given.

y - y1 = m(x - x1)

y - 1/3 = 3/4(x - 4)

Answer:

y - 1/3 = 3/4(x - 4)

6 0
3 years ago
timothyis re-arranging his marbles collection. he has 5 identical blue marbles, five green and 3 black. he can fit excactly 5 ma
Lerok [7]
-- He must have at least one of each color in the case, so the first 3 of the 5 marbles in the case are  blue-green-black.

Now the rest of the collection consists of

                    4 blue
                    4 green
                    2 black

and there's space for 2 more marbles in the case.

So the question really asks:  "In how many ways can 2 marbles
be selected from 4 blue ones, 4 green ones, and 2 black ones ?"

-- Well, there are 10 marbles all together. 
So the first one chosen can be any one of the 10,
       and for each of those,
the second one can be any one of the remaining 9 .

Total number of ways to pick 2 out of the 10  =  (10 x 9)  =  90 ways.

-- BUT ... there are not nearly that many different combinations
to wind up with in the case.    

The first of the two picks can be any one of the 3 colors,
             and for each of those,
the second pick can also be any one of the 3 colors. 

So there are actually only 9 distinguishable ways (ways that
you can tell apart) to pick the last two marbles.
6 0
3 years ago
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