<em>BD</em> = 56
Step-by-step explanation:
Step 1: In rectangle, the diagonals are congruent and bisect each other.
So, <em>AC</em> = <em>BD</em>
⇒<em>AG</em> + <em>GC</em> = <em>BG</em> + <em>GD</em>
⇒<em>AG</em> + <em>AG</em> = <em>GD</em> + <em>GD</em>
⇒ 2<em>AG</em> = 2<em>GD</em>
⇒<em>AG</em> = <em>GD</em>
⇒ –7<em>j </em>+ 7 = 5<em>j</em> + 43
⇒–7<em>j</em> – 5<em>j</em> = 43 – 7
⇒–12<em>j</em> = 36
⇒<em>j</em> = –3
Step 2: <em>BD</em> = 2<em>DG</em>
<em>BD</em> = 2(5<em>j</em> + 43)
= 2(5 (–3) + 43)
= 2(–15 + 43)
= 2 × 28
= 56
Hence, <em>BD</em> = 56.
Answer:
either b or c but i think its b
Step-by-step explanation:
Answer:
0.0918
Step-by-step explanation:
We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The mean and standard deviation of average spending of sample size 25 are
μxbar=μ=95.25
σxbar=σ/√n=27.32/√25=27.32/5=5.464.
So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The z-score associated with average spending $102.5
Z=[Xbar-μxbar]/σxbar
Z=[102.5-95.25]/5.464
Z=7.25/5.464
Z=1.3269=1.33
We have to find P(Xbar>102.5).
P(Xbar>102.5)=P(Z>1.33)
P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)
P(Xbar>102.5)=0.5-0.4082
P(Xbar>102.5)=0.0918.
Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.
I think you have to do step by step and so you have the input and output chart so 32 | 20 find out the pattern its going by like what +,/,*,- would get you to 20 the best I can help
