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Ber [7]
4 years ago
6

1. What is the mean change in the forecasted high temperatures over the next 7 days? Remember, this can be found by averaging th

e values in the Difference column for the high temperatures. Show your work and steps. If your answer is not an integer, explain what two integers your answer is between. 2. What is the mean change in the forecasted low temperatures over the next 7 days? Remember, this can be found by averaging the values in the Difference column for the low temperatures. If your answer is not an integer, explain what two integers your answer is between.

Mathematics
1 answer:
Schach [20]4 years ago
4 0

Answer:

The table is missing in the question. The table is provided below.

Step-by-step explanation:

1. In the table, it is given the difference of high temperature. They are :

-7, -7, 2, 5, -3, -2

Now adding the differences of high temperatures and taking out its average.

-7 + (-7) + 2 + 5 + (-3) + (-2)= -12

Average : $-\frac{12}{6}=-2$

Thus the answer is an integer.

2.  In the table, it is given the difference of high temperature. They are :

0, -3, -7, -1, 1, 0

Now adding the differences of high temperatures and taking out its average.

0 + (-3) + (-7 )+ (-1) + 1 + 0= -10

Average : $-\frac{10}{6} = -1.66$

Thus, the answer is not an integer. The answer lies between the integers -2 to -1.

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Yanira is 3 years older than tim and twice as old as hannah. tim is 2 years older than hannah. how old are yanira, tim, and hann
Sati [7]
Y = T+3 (i)
T = H+2  (ii)
Y = 2*H  (iiii)

T = H+2  therefore H = T-2

Substitute H=T-2 into (iii)
Y=2*H
Y=2*(T-2)
Y=2T-4

Now substitute that Y into (i)
(i) says Y = T+3
2T-4=T+3
2T-T-4=3
T-4=3
T=3+4
T=7

Then from (i) Y=T+3 = 7+3 =10
Y=10

And from (iii) Y=2*H
If Y = 10, then H = 5

4 0
3 years ago
Which is the equation of a hyperbola with directrices at x = ±3 and foci at (4, 0) and (−4, 0)?​
Alex Ar [27]

Answer:

\frac{x^2}{12}-\frac{y^2}{4}=1

Step-by-step explanation:

Since our foci are located on the x-axis, then our major axis is going to be the horizontal transverse axis of the hyperbola:

<u>Formula for hyperbola with horizontal transverse axis centered at origin</u>

  • <u />\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
  • Directrices -> x=\pm\frac{a^2}{c}
  • Foci -> (\pm c,0) where a^2+b^2=c^2
  • a>b

Since we are given our directrices of x=\pm3 and foci of (\pm4,0), then we can set up the directrices equation to solve for a^2:

x=\pm\frac{a^2}{c}\\ \\\pm3=\pm\frac{a^2}{4}\\ \\12=a^2

Now we can determine b^2 to complete our equation for the hyperbola:

a^2+b^2=c^2\\\\12+b^2=4^2\\\\12+b^2=16\\\\b^2=4

Therefore, our equation for our hyperbola is \frac{x^2}{12}-\frac{y^2}{4}=1

5 0
2 years ago
Does anyone know how to do the problem
salantis [7]
8x^3 + 2x^2 - 18x -12
6 0
3 years ago
Help with question pls I don't understand​
Mazyrski [523]

Answer:

15x^3y+6x^2y^2+9xy^3

Step-by-step explanation:

3xy ( 5x^2 +2xy +3y^2)

Distribute the 3xy to all terms in the parentheses

3xy*5x^2+3xy*2xy+3xy*3y^2

15x^3y+6x^2y^2+9xy^3

6 0
3 years ago
Read 2 more answers
Explain when the median of a data set is a better<br> measure of center than the mean.
Delvig [45]

Answer:

The median of a data set is better when you have a term or terms that are not close to the other terms

Step-by-step explanation:

For example:

Say you have the data set

1, 15, 17, 18, 22, 84

The median of these terms would be 17.5

       (it is the exact center of the data group)

17 + 18 = 35

35/2 = 17.5

The mean of these terms would be 26.17  

    (this number is not close to the center because the numbers 1 and 84

      are not close enough to the other terms)

1 + 15 + 17 + 18 + 22 + 84 = 157

157/6 = 26.17

7 0
3 years ago
Read 2 more answers
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