How do you solve 44 ÷ 0.5
2 answers:
You might be interested in
<h2>(1)</h2><h2> =(a+b)(3c-d)</h2><h2> =a(3c-d)+b(3c-d)</h2><h2> =3ac-ad+3bc-bd</h2><h2>(2)</h2><h2> =(a-b)(c+2d)</h2><h2> =a(c+2d)-b(c+2d)</h2><h2> =ac+2ad-bc-2bd</h2><h2>(3)</h2><h2> =(a-b)(c-2d)</h2><h2> =a(c-2d)-b(c-2d)</h2><h2> =ac-2ad-bc+2bd</h2><h2>(4)</h2><h2> =(2a+b)(c-3d)</h2><h2> =2a(c-3d)+b(c-3d)</h2><h2> =2ac-6ad+bc-3bd</h2>
Answer:
1) Pr(all three) = 0.064
2) Pr(none) = 0.216
3) Independent event
Step-by-step explanation:
Let Probability of homes constructed in the Quail Creek area with a security system = Pr (security)
Where Pr = probability
This is a binomial distribution problem. There ate only two outcomes in this distribution: a success and a failure
Where p = success, q = failure
For n trials,
Pr(X = x) = n!/[(n-x)!x!] × p^x × q^(n-x)
Pr (security) = 40% = 0.4
p = 0.4
q = 1-p = 1-0.4 = 0.6
Numbers selected at random = n = 3
See attachment for more details on the workings.
1) Probability all three of the selected homes have a security system = Pr(all three)
Pr(all three) = 1× (0.4)³ (0.6)^0 = 0.4³
= 0.4×0.4×0.4
Pr(all three) = 0.064
2) Probability none of the three selected homes have a security system
= Pr(none)
Pr(none) = 1 × (0.4)^0 × (0.6)³
= 0.6³ = 0.6×0.6×0.6
Pr(none) = 0.216
3) The events were assumed to be independent as the selection of one house doesn't affect the outcome of the selection of another.
