Answer:
6x-11y=-13
Step-by-step explanation:
(x1,y1)=(3/2,2) and (x2,y2)=(-4,-1)
y-y1= y2-y1/x2-x1 (x-x1)
y-2=-3/-11/2(x-3/2)
=6/11(x-3/2)
11(y-2)=6x-9
11y-22=6x-9
6x-11y=9-22
6x-11y=-13
<h3>Given:</h3>
- P= $50,000
- R= 10%
- T= 5 years
<h3>Note that:</h3>
- P= Principal amount
- R= Rate of interest
- T= Time period
<h3>Solution:</h3>

Let's substitute according to the formula.

<em>A=</em><em> </em><em>$80525.5</em>
Now, we can find the interest paid

We'll have to deduct the total amount from the principal amount.
Let's substitute according to the formula.

<em>I=</em><em> </em><em>$30525.5</em>
<u>Hence</u><u>,</u><u> </u><u>the</u><u> </u><u>total</u><u> </u><u>amount</u><u> </u><u>paid</u><u> </u><u>after</u><u> </u><u>5</u><u> </u><u>years</u><u> </u><u>is</u><u> </u><u>$</u><u>80525.5</u><u> </u><u>and</u><u> </u><u>$</u><u>30525.5</u><u> </u><u>was</u><u> </u><u>paid</u><u> </u><u>as</u><u> </u><u>interest</u><u>.</u>
Answer:
P ( -1 < Z < 1 ) = 68%
Step-by-step explanation:
Given:-
- The given parameters for standardized test scores that follows normal distribution have mean (u) and standard deviation (s.d) :
u = 67.2
s.d = 4.6
- The random variable (X) that denotes standardized test scores following normal distribution:
X~ N ( 67.2 , 4.6^2 )
Find:-
What percent of the data fell between 62.6 and 71.8?
Solution:-
- We will first compute the Z-value for the given points 62.6 and 71.8:
P ( 62.6 < X < 71.8 )
P ( (62.6 - 67.2) / 4.6 < Z < (71.8 - 67.2) / 4.6 )
P ( -1 < Z < 1 )
- Using the The Empirical Rule or 68-95-99.7%. We need to find the percent of data that lies within 1 standard about mean value:
P ( -1 < Z < 1 ) = 68%
P ( -2 < Z < 2 ) = 95%
P ( -3 < Z < 3 ) = 99.7%
Answer:
7/10
Step-by-step explanation:
2/5 multiply the denominator by 2 which will make it 10. Then multiply the numerator by 2 which will make it 4. So it will be 4/10. Then add 4/10 to 3/0 which makes 7/10.