Answer:
Step-by-step explanation:
Hypotenuse² =base² + altitude²
(3x + 4)² = (2x + 1)² + (3x)²
{Use (a+b)² = a² + 2ab + b²}
(3x)² + 2*3x+4 + 4² = (2x)² + 2*2x*1 + 1² + 9x²
9x² + 24x + 16 = 4x² + 4x + 1 + 9x²
9x² + 24x + 16 = 13x² + 4x + 1
0 = 13x² + 4x + 1 - 9x² - 24x - 16
13x² - 9x² + 4x - 24x +1 - 16 = 0
4x² - 20x - 15 = 0
a = 4 ; b =-20 ; c = -15
D = b² - 4ac = (-20)² - 4*4*(-15) = 400 + 240 = 640
√D = √640 = 25.30
x = 5 .66 ; x = -0.66
Ignore x = -0.66 as length of a side cannot be negative
Answer : x = 5.66
Answer:
C) 960
Step-by-step explanation:
Let's expand the factorial.
n! = 1*2*3*4*...n
6! = 1*2*3*4*5*6
3!= 1*2*3
The given expression
=
Canceling out the common terms, we get
=
= 8.4.5.6 [here . represents multiplication]
= 960
Answer: C) 960
Thank you.
Answer:
(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.
(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.
(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.
Step-by-step explanation:
We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.
Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.
Let X = <u><em>back-to-college family spending on electronics</em></u>
SO, X ~ Normal()
The z score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = population mean family spending = $237
= standard deviation = $54
(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)
P(X < $150) = P( <
) = P(Z < -1.61) = 1 - P(Z
1.61)
= 1 - 0.9463 = <u>0.0537</u>
The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.
(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)
P(X > $390) = P( >
) = P(Z > 2.83) = 1 - P(Z
2.83)
= 1 - 0.9977 = <u>0.0023</u>
The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.
(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)
P($120 < X < $175) = P(X < $175) - P(X $120)
P(X < $175) = P( <
) = P(Z < -1.15) = 1 - P(Z
1.15)
= 1 - 0.8749 = 0.1251
P(X < $120) = P( <
) = P(Z < -2.17) = 1 - P(Z
2.17)
= 1 - 0.9850 = 0.015
The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.
Therefore, P($120 < X < $175) = 0.1251 - 0.015 = <u>0.1101</u>
The distributive property lets you expand the product as