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3 years ago

Find the value of x so f(x)=13f(x)=3x+4

Mathematics

1 answer:

svetlana [45]3 years ago

3 0

Answer:

Step-by-step explanation:

3(13)+4= 43

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Instruction

slavikrds [6]

<h3>Answer: Choice C. (-x, -y)</h3>

Explanation:

Focus on one point, such as A = (1,2). Note how it moves to point A ' = (-1, -2). Both x and y coordinates have been flipped from positive to negative. The rule therefore is $(x,y) \to (-x,-y)$ . This describes a 180 degree rotation (either clockwise or counterclockwise, it doesn't matter). Points B and C follow the same idea.

Side note: Lines AA', BB' and CC' all go through the origin (0,0).

8 0

3 years ago

1462 divided by 58<br><br> To mixed number

GaryK [48]

Ur answer is 25/6/29 plz give brainliest

6 0

3 years ago

Alexandra has a stand in the marketplace where she sells her cumin for 7.20 per kilogram. If she sells 90 kilograms of cumin her

Drupady [299]

Answer:

$234

Step-by-step explanation:

First we need to define profits. Profits are Income minus Expenses:

P = I - E

We know profits are $414, so:

414 = I - E

We also can calculate income, as it is equal to price by the sales:

I = p*Q

Here she sold 90 kgs at $7.20 b kg. So:

I = p*Q = 7.20 * 90 = 648

So, replacing in profits equation:

414 = I - E

414 = 648 - E

If we sum E in both sides:

414 + E = 648 - E + E = 648

414 + E = 648

Now, subtracting 414 in both sides:

414 + E - 414 = 648 - 414

E = 234

So, her expenses are $234

8 0

3 years ago

What the result of √[(-2a)^ 2]

Tanya [424]

If you take the square root of a number squared number then they cancel each other out and the number stays the same i.e. √[(4)^2] would equal 4.

In this problem the square root and numbered squared cancel out to leave the problem as -2a.

The solution of this problem is -2a

3 0

3 years ago

Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus

quester [9]

Hi!

To compare this two sets of data, you need to use a t-student test:

You have the following data:

-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph

-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

$t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510$

To calculate the degrees of freedom you need to use the following equation:

$df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89$ ≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.

5 0

3 years ago