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serious [3.7K]
3 years ago
12

Slope intercept form

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0
The slope intercept form is y = me+b
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Pls help!!! Which list of numbers is ordered from greatest to least?
Anettt [7]
B

It’s the only answer that begins with a positive number
6 0
3 years ago
Casey was twice as old as his sister 3 years ago. Now he is 5 years older than his sister. How old is Casey?
Margarita [4]
Let x is Casey's sister age.
<span>3 years ago: 
</span><span>Casey was twice as old as his sister: 2a - 3
Now: 
</span><span>he is 5 years older than his sister: a + 5

2a - 3 = a + 5
subtract a to both sides
2a - 3 -a = a + 5 - a
simplify
a - 3 = 5
add 3 to both sides
a - 3 + 3 = 5 + 3
simplify
a = 8

His sister age is 8 years old now.
</span>Now: 
he is 5 years older than his sister: a + 5 = 8 + 5 = 13
<span>
proof: 
so 3 years ago,
his sister age: 8 - 3 = 5
Casey age 3 years ago:
13 - 3 = 10 (</span>C<span>asey was twice as old as his sister 3 years ago</span>)
<span>
Answer:
Casey is 13 years old now.


</span>
4 0
4 years ago
Read 2 more answers
The heart rate of 10 adults is measured and the results are 83, 87, 90, 92, 93, 100, 104, 111, 115, 121. Find the interquartile
zloy xaker [14]

The interquartile range of the data set is 21.

Given that the heart rate of 10 adults is measured and the results are 83, 87, 90, 92, 93, 100, 104, 111, 115, 121.

The interquartile range shows the extent of the middle half of the distribution. Quartiles segment any distribution, ordered from low to high, into four equal parts. The interquartile range (IQR) contains the second and third interquartile ranges, the central half of the dataset.

The value of quartile 1 is Q₁=90

The value of quartile 3 is Q₃=111

So, the interquartile range is

IQR=Q₃-Q₁

IQR=111-90

IQR=21

Hence, the interquartile range of the data set 83, 87, 90, 92, 93, 100, 104, 111, 115, 121 is 21.

Learn more about the interquartile range from here brainly.com/qustion/18723655

#SPJ4

6 0
2 years ago
The half-life of radioactive actinium (^227 Ac) is 22 years. How much will remain after 15 years? Find R first.
Llana [10]
Use the formula below.

Let A_0 = 1 because you just want to know the percentage left. Usually that's the initial amount.

t = 15, the number of years past. n = 22, the half-life.

A=A_0(0.5)^{\frac{t}{n}} \\  \\ A=0.5^{\frac{15}{22}} \\  \\ A=0.62 = 62\%
5 0
3 years ago
Total number of chocolate boxes that can be produced: x+y (&lt;,&lt;=,&gt;,&gt;=) ___
Illusion [34]

Answer:

Step-by-step explanation:

The idea here is to create lines according to the constraints we were given, graph the lines (which are actually inequalities), and then shade in the region that satisfies the inequality. Let's start at the beginning of the problem and we'll get our lines (inequalities) written.

The total number of boxes that can be produced according to the constraints is 800, so the inequality for that is

x + y ≤ 800 and solving for y:

y ≤ 800 - x

Another constraint on the white chocolate is that it has to be less than or equal to 200 boxes, so:

y ≤ 200

The max number of white chocolate boxes is half the number of milk chocolate, so:

y ≤ (1/2)x

The min number of milk chocolate boxes produced is:

x ≥ 0 and

The min number of white chocolate boxes produced is:

y ≥ 0 (This means that it is a possibility of making 0 milk chocolate boxes and all white chocolate boxes OR there is a possibility of making 0 white chocolate boxes and all milk chocolate boxes)

The production equation (which is used later) is:

2.25x + 2.50y (you make a profit of $2.25 on every milk chocolate box you sell and profit of $2.50 on every white chocolate box you sell).

The bold equations are the ones that need to be graphed (see graph below). Where those 3 lines intersect are the vertices of feasible region:

(0, 0), (400, 200), (600, 200), (800, 0).

Then take each x and y value from a coordinate and plug it into the profit equation (we don't need to use (0, 0)) starting with x = 400 and y = 200:

2.25(400) + 2.5(200) = $1400

Now using x = 600 and y = 200:

2.25(600) + 2.5(200) = $1850

Now using x = 800 and y = 0:

2.25(800) + 2.5(0) = $1800

So our max profit as seen by the evaluations is $1850, and that occurs when we sell 600 boxes of milk chocolate and 200 boxes of white chocolate.

6 0
3 years ago
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