The factors of 50a³ are 1, 2, 5, 10, 25, 50,
and their products with a, a² and a³ .
The factors of 10a² are 1, 2, 5, 10,
and their products with 'a' and a² .
Their common factors are 1, 2, 5, 10,
and their products with 'a' and a².
Their greatest common factor is 10a² .
(Another way to spot it, easily, is to remember this helpful factoid:
If the smaller number is a factor of the larger number,
then the smaller number is their greatest common factor.
Using the greatest common factor, then . . .
50a³ + 10a² = 10a²(5a + 1) .
Answer is a, 57.49.
17.42-12.60-9.62= $-4.80+62.29=57.49
To answer this item, we make use of the equation derive from the Pythagorean theorem for right triangles which states that the square of the hypotenuse (the longest side) is equal to the sum of the squares of the two shorter sides. If we let x be the measure of both the shorter sides (we call this as legs), we have,
(14 in)² = (x²) + (x²)
Simplifying the equation,
196 in² = 2x²
Divide both sides of the equation by 2,
98 in² = x²
To get the value of x, we get the square root of both sides of the equation,
x = sqrt (98) = 7√2 inches
Hence, the measure of each leg of the right triangle is 7√2 inches or approximately 9.9 inches.
Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
Group 1:
μ1 = 59.7
s1 = 2.8
n1 = sample size = 12
Group 2:
μ2 = 64.7
s2 = 8.3
n2 = sample size = 15
α = 0.1
Assume normal distribution and equ sample variance
A.)
Null and alternative hypothesis
Null : μ1 = μ2
Alternative : μ1 < μ2
B.)
USing the t test
Test statistic :
t = (m1 - m2) / S(√1/n1 + 1/n2)
S = √(((n1 - 1)s²1 + (n2 - 1)s²2) / (n1 + n2 - 2))
S = √(((12 - 1)2.8^2 + (15 - 1)8.3^2) / (12 + 15 - 2))
S = 6.4829005
t = (59.7 - 64.7) / 6.4829005(√1/12 + 1/15)
t = - 5 / 2.5108165
tstat = −1.991384
Decision rule :
If tstat < - tα, (n1+n2-2) ; reject the Null
tstat < t0.1,25
From t table :
-t0.1, 25 = - 1.3163
tstat = - 1.9913
-1.9913 < - 1.3163 ; Hence reject the Null
Answer:
59%
Step-by-step explanation:
simply, it would be 11+2/22 = 59%
I assume there are no student that passed 0 tests
