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Yanka [14]
2 years ago
11

I NEED HELP NOW What is the equation in slope-intercept form of the linear function represented by the table? x y –6 –18 –1 –8 4

2 9 12 y = negative 2 x minus 6 y = negative 2 x + 6 y = 2 x minus 6 y = 2 x + 6
Mathematics
2 answers:
Bogdan [553]2 years ago
7 0

Answer:

its was d for me

Step-by-step explanation:

correct if im wrong

sergiy2304 [10]2 years ago
5 0

Answer:

it woupld be the lterr if 7 and 65 and juj hyhy st

Step-by-step explanation:

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A teacher gives a test to a large group of students. The results are closely approximated by a normal curve. The mean is 74​, wi
Alinara [238K]

Answer:

Step-by-step explanation:

Given that a teacher  gives a test to a large group of students. The results are closely approximated by a normal curve

mu =74 and sigma =8

A grade starts from 100-8 = 92nd percentile

Z score for 92nd percentile = 1.405

X score = 74+8(1.405) = 85.24

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

B cut off is to next 16%

Hence C would start for scores below 100-(8+16) = 76%

76th percentile = 0.705*8+74 =79.64

6 0
2 years ago
A computer store decides to increase the prices of all the items it sells by 15%. The store manager uses matrices to prepare the
Nitella [24]
The manager could perform scalar multiplication on Matrix A, using the scalar 1.15.

Increasing the price by 15% would mean we are taking 100% of the value + another 15%; 100+15 = 115%; 115% = 115/100 = 1.15.

Multiplying every value in Matrix A by 1.15 will give the price raised by 15%.
4 0
2 years ago
You ask 150 people about their pets. The results show that 9/25 of the people own a dog. Of the people who own a dog. 1/6 of the
pav-90 [236]

Answer: 9

Step-by-step explanation:

In order to find the people out of the 150, we first multiply 9/25 and 150 together. This will give us the people who DO own a dog. This brings us to 54.

Now, if we multiply 54 by 1/6 we get the answer of 9, because 1/6 of the dog owners own a cat as well. We have now reached the desired amount.

6 0
1 year ago
Which equation represents the line that passes through the points (6, –3) and (–4, –9)?
Delvig [45]
(6,-3)(-4,-9)
slope(m) = (-9 -(- 3) / (-4 - 6) = (-9 + 3) / -10 = -6/-10 = 3/5

y - y1 = m(x - x1)
slope(m) = 3/5
(6,-3)....x1 = 6 and y1 = -3
now we sub
y - (-3) = 3/5(x - 6) =
y + 3 = 3/5(x - 6) <==
8 0
2 years ago
Read 2 more answers
An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "
FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

\large X_{sentences} the mean of the “sentences” group

\large S_{sentences} the standard deviation of the “sentences” group

\large X_{intentional} the mean of the “intentional” group

\large S_{intentional} the standard deviation of the “intentional” group

Then, we can calculate by using the computer

\large X_{sentences}=28.75  

\large S_{sentences}=3.53553

\large X_{intentional}=31.625

\large S_{intentional}=1.40788

\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875

The <em>standard error of the difference (of the means)</em> for a sample of size 8 is calculated with the formula

\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}

So, the standard error of the difference is

\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546

<em>In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of  95% for the difference of the means corresponding to a level of significance of 0.05 (5%). </em>

<em>If this interval contains the zero, we can say there is no significant difference. </em>

<em>Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples. </em>

We get the \large t_{0.05} which is a value of t such that the area under the Student's t distribution  outside the interval \large [-t_{0.05}, +t_{0.05}] is less than 0.05.

That value can be obtained either by using a table or the computer and is found to be

\large t_{0.05}=2.365

Now we can compute our confidence interval

\large (X_{sentences}-X_{intentional}) \pm t_{0.05}*(standard \;error)=-2.875\pm 2.365*1.34546

and the confidence interval is

[-6.057, 0.307]

Since the interval does contain the zero, we can say there is no significant difference in these samples.

6 0
2 years ago
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