Answer:
THere will be 8 heads and 2 tails
Step-by-step explanation:
I don't know
a+b+c=9 and a2+b2+c2=35 then by using identity
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca,we get the answer 9x9=35+2(ab+bc+ca)
2(ab+bc+ca)=9x9-35
2(ab+bc+ca) =81-35then we get ab+bc+ca=46/2 so ab+bc+ca=23
To solve this problem you must apply a theorem called: Intersecting secant theorem. The proccedure is shown below:
1) For the figure shown on the left:
(5)(x+5)=(6)(4+6)
5x+25=60
5x=60-25
x=35/5
x=7
2) For the figure shown on the right:
(3)(3+5)=(4)(x+4)
24=4x+16
24-16=4x
8=4x
8/4=x
x=2
You are right to circle both 10 or 12. Neither can be solved without knowing something about the x values in 10 and the x value of the smaller base in 12.
Answer:
We reject H₀
we accept Hₐ seeds in the packet would germinate smaller than 93%
Step-by-step explanation:
Test of proportions
One tail-test (left side)
93 % = 0.93
p₀ = 0,93
1.- Hypothesis
<h3>
H₀ ⇒ null hypothesis p₀ = 0.93</h3><h3>
Hₐ ⇒ Alternative hypothesis p = 0.875</h3><h3>
2.-Confidence interval 95 %</h3><h3>
α = 0,05 </h3><h3>
and </h3><h3>
z(c) = - 1.64</h3><h3>
3.- Compute z(s)</h3><h3>
z(s) = (p - p₀)/√(p₀*q₀)/n z(s) = (0.875-0.93)/√0.93*0.07)200</h3><h3>
z(s) = - 0,055/ √0.0003255</h3><h3>
z(s) = - 0.055/ 0.018</h3><h3>
z(s) = - 3,06</h3><h3>
4.-Compere z(c) and z(s)</h3><h3>
z(s) < z(c) -3.06 < -1.64</h3><h3>
z(s) is in rejection region, we reject H₀</h3>
