How do i graph y= 2x

What is the solution set for the equation <img src="https://tex.z-dn.net/?f=%5Csqrt%7B56-x%3Dx%7D" id="TexFormula1" title="\sqrt

Juliette [100K]

Answer:

A.

{7}

Step-by-step explanation:

hello

sqrt(56-7) =sqrt(49)=7

7 0

2 years ago

Sally has been measuring the tree in her yard for a science project. The tree has grown 12 of an inch each month and grown a tot

hammer [34]

Answer:

Sally has been measuring the tree for 4 months

Step-by-step explanation:

The correct question is as follows;

Sally has been measuring the tree in her yard for a science project. The tree has grown 1/2 of an inch each month and grown a total of 2 inches taller. How many months has Sally been measuring this tree?

.........................................................................................

Solution;

The tree has grown 1/2 of an inch each month and has grown a total of 2 inch taller.

We now need to know the number of months in which the measuring has been taking place.

To know the number of months, we simply need to divide the change in height which is 2 inch by the individual month growth rate which is 1/2 inch

So mathematically, that would be 2/1/2 = 2 * 2 = 4 months

3 0

2 years ago

An economy pack of highlighters contains 12 yellow, 6 blue, 4 green, and 3 orange highlighters. An experiment consists of random

Serhud [2]

Answer:

The probability of choosing a blue or yellow highlighter when one yellow highlighter is picked is 17/25.

Step-by-step explanation:

Here, according to the question:

Total number of yellow highlighters = 12

Total number of blue highlighters = 6

Total number of green highlighters = 4

Total number of orange highlighters = 3

So, the total number of highlighters = 12 + 6 + 4 + 3 = 25

Let E1 : Event of selecting a yellow highlighter.

P(E1) = $\frac{\textrm{Total number of yellow highlighters}}{\textrm{Total Highlighters}}$ = $\frac{12}{25}$

Let E2 : Event of selecting blue or yellow highlighter ONCE yellow highlighter is selected.

So, the total yellow highlighters left after selecting one = 12 -1 = 11

Also, the number of blue highlighter = 6

So, the TOTAL FAVORABLE options to E2 = 6+ 11 = 17

P(E2) =

$\frac{\textrm{Total number of yellow + Blue highlighters}}{\textrm{Total Highlighters}}$ = $\frac{17}{25}$

Hence, the probability that a blue or yellow highlighter is selected given that a yellow highlighter is selected is 17/25.

7 0

3 years ago

Greg had $17 more than Fred, and together they had just enough money to by their mom a present that costs $85. How much money di

babunello [35]

<u>Define x:</u>

Let Fred had x dollar.

Fred = x
Greg = x + 17

<u>Solve x :</u>

x + x + 17 = 85
2x + 17 = 85
2x = 68
x = 34

<u>Find the amount each of them had:</u>

Fred = x = $34
Greg = x + 17 = 34 + 17 = $51

Answer: Greg had $51

3 0

3 years ago

Read 2 more answers

The final scores from the last five games of two basketball teams are given. Jumping Jackrabbits: 70, 65, 72, 80, 73 Leaping Liz

scoundrel [369]

Using the Mean Absolute Deviation(MAD) concept, it is found that:

a) The Jumping Jackrabbits have a MAD of 3.6 and the Leaping Lizards of 6.8.

b) The MAD is lower for the Jumping Jackrabbits then for the Leaping Lizards, hence their scores are more consistent.

<h3>What is the mean absolute deviation of a data-set?</h3>

The mean of a data-set is given by the sum of all observations divided by the number of observations.

The mean absolute deviation(MAD) of a data-set is the sum of the absolute value of the difference between each observation and the mean, divided by the number of observations.

The mean absolute deviation represents the average by which the values differ from the mean.

Item a:

For the Jumping Jackrabbits, the mean and MAD are given by:

M = (70 + 65 + 72 + 80 + 73)/5 = 72

MAD = (|70 - 72| + |65 - 72| + |72 - 72| + |80 - 72| + |73 - 72|)/5 = 3.6.

For the Leaping Lizards, the mean and the MAD are given by:

M = (61 + 47 + 70 + 63 + 54)/5 = 59

MAD = (|61 - 59| + |47 - 59| + |70 - 59| + |63 - 59| + |54 - 59|)/5 = 6.8.

The Jumping Jackrabbits have a MAD of 3.6 and the Leaping Lizards of 6.8.

Item b:

The MAD is lower for the Jumping Jackrabbits then for the Leaping Lizards, hence their scores are more consistent.

More can be learned about the Mean Absolute Deviation(MAD) concept at brainly.com/question/3250070

#SPJ1

4 0

2 years ago