How do you graph x-2y=4 And 2x+y=-7 system of equations

A school district is considering moving the start of the school day at Groveland High from 8:00 a.m. to 9:00 a.m. to allow stude

DaniilM [7]

Answer:

The answer is " $\bold{(7.1-8.3) \pm 1.665 \sqrt{\frac{(1.7)^2}{50}+\frac{(1.9)^2}{39}} }\\\\$ "

Step-by-step explanation:

Given values:

$\bar{x_1}=7.1\\\\\bar{x_2}=8.3\\\\s_1=1.7\\\\s_2=1.9\\\\n_1=50\\\\n_2=39$

Using formula:

$\to (\bar{x_1} -\bar{x_2}) \pm t \sqrt{\frac{(s_1)^2}{n_1}+\frac{(s_1)^2}{n_2}} \\\\$

Put the values in the above formula:

$\to (7.1-8.3) \pm 1.665 \sqrt{\frac{(1.7)^2}{50}+\frac{(1.9)^2}{39}} \\\\$

4 0

3 years ago

The amount of daily time that teenagers spend on a brand A cell phone is normally distributed with a given mean mc027-1.jpg = 2.

ivolga24 [154]

<span>16% is the answer Fam.....................................</span>

4 0

3 years ago

Read 2 more answers

In a game of chance, the probability of winning a $50 is 40% and the probability of losing a $50 prize is 60%. what is the expec

melamori03 [73]

To calculate expected value,
We can add up the probability of all possible events.
The 2 possible events are either winning $50 or losing the prize, which is $ - 50
Expected value =
50 x 40 % + (-50) x 60%
= $-10
So the answer is $-10

4 0

3 years ago

If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t

sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, $6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296$

This implies that $6^{4}$ exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, $6^{3}=216$ and $216\times 5=1080, 216\times 4=864$ . This implies that $216\times 5$ exceeds 1052 and thus there can be at most 4 groups of $6^{3}$ .

Then,

$1052-4\times6^{3}$

$1052-4\times216$

$1052-864$

$188$

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, $6^{2}=36$ and $36\times 5=180, 36\times 6=216$ . This implies that $36\times 6$ exceeds 188 and thus there can be at most 5 groups of $6^{2}$ .

Then,

$188-5\times6^{2}$

$188-5\times36$

$188-180$

$8$

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, $6^{1}=6$ and $6\times 1=6, 6\times 2=12$ . This implies that $6\times 2$ exceeds 8 and thus there can be at most 1 group of $6^{1}$ .

Then,

$8-1\times6^{1}$

$8-1\times6$

$8-6$

$2$

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, $6^{0}=1$ and $1\times 2=2$ . This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

Then,

$1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}$