Answer:
Mean = 0.2(100) = 20
Variance = 0.2×0.8×100 = 16
Standard deviation = 4
Yes, np > 5 & nq > 5
Answer:
(-1,6)
Step-by-step explanation:
Ultimately, you want the <em>variable</em> p on one side of the equals sign, and the <em>value</em> of p on the other.
First, get rid of the parenthesis by <em>distributing.
</em>There is an invisible -1 in front of the first and last set of parenthesis, and a one times any number is the number itself, so you are essentially distributing the negative sign.
<em>
</em>So you have: 2-4p-3 = -3p-6-2p-3
<em>
</em>Next, <em>Combine like terms.
</em>
Can you take it from there?<em>
</em>
Answer:
<h3>1. SAS</h3><h3>2. SSS</h3><h3>3. AAS</h3><h3>4. AAS</h3><h3>5. AAS</h3><h3>6. HL</h3><h3>7. SAS</h3><h3>8. SAS</h3><h3>9. SSS</h3>
Answer:
An expression for Francisco's age in terms of a is (2a+3)
An expression for Harrison's age in terms of a is (3a-2)
An expression for the sum of all three ages, in terms of a is (6a+1)
Step-by-step explanation:
Let
a -----> Jenna's age
b -----> Francisco's age
c -----> Harrison's age
we know that
Francisco's age is two times Jenna's age plus three years
b=2a+3 -----> equation A
Harrison's age is three times Jenna's age minus two years
c=3a-2 ----> equation B
The sum of all three ages, in terms of a is
(a+b+c) -----> equation C
substitute equation A and equation B in equation C
a+(2a+3)+(3a-2)
Combine like terms
(a+2a+3a)+(3-2)
6a+1
therefore
An expression for Francisco's age in terms of a is (2a+3)
An expression for Harrison's age in terms of a is (3a-2)
An expression for the sum of all three ages, in terms of a is (6a+1)