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34kurt
3 years ago
7

I don't understand how to do this problem. help ? ( number 13)

Mathematics
1 answer:
OverLord2011 [107]3 years ago
7 0
5x- 10 = 20
5x -10 +10 = 20 +10
5x = 30
5x/5 =30/5
x = 6
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1.80

Step-by-step explanation:

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D =m/v
yKpoI14uk [10]

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Step-by-step explanation:

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Which could be the length of BC?<br><br> 11 ft<br><br> 13 ft<br><br> 15 ft<br><br> 18 ft
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6 0
3 years ago
Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Ro
puteri [66]

Answer:

(a) P(0 ≤ Z ≤ 2.87)=0.498

(b) P(0 ≤ Z ≤ 2)=0.477

(c) P(−2.20 ≤ Z ≤ 0)=0.486

(d) P(−2.20 ≤ Z ≤ 2.20)=0.972

(e) P(Z ≤ 1.01)=0.844

(f) P(−1.95 ≤ Z)=0.974

(g) P(−1.20 ≤ Z ≤ 2.00)=0.862

(h) P(1.01 ≤ Z ≤ 2.50)=0.150

(i) P(1.20 ≤ Z)=0.115

(j) P(|Z| ≤ 2.50)=0.988

Step-by-step explanation:

(a) P(0 ≤ Z ≤ 2.87)

In this case, this is equal to the difference between P(z<2.87) and P(z<0). The last term is substracting because is the area under the curve that is included in P(z<2.87) but does not correspond because the other condition is that z>0.

P(0 \leq z \leq 2.87)= P(z

(b) P(0 ≤ Z ≤ 2)

This is the same case as point a.

P(0 \leq z \leq 2)= P(z

(c) P(−2.20 ≤ Z ≤ 0)

This is the same case as point a.

P(-2.2 \leq z \leq 0)= P(z

(d) P(−2.20 ≤ Z ≤ 2.20)

This is the same case as point a.

P(-2.2 \leq z \leq 2.2)= P(z

(e) P(Z ≤ 1.01)

This can be calculated simply as the area under the curve for z from -infinity to z=1.01.

P(z\leq1.01)=0.844

(f) P(−1.95 ≤ Z)

This is best expressed as P(z≥-1.95), and is calculated as the area under the curve that goes from z=-1.95 to infininity.

It also can be calculated, thanks to the symmetry in z=0 of the standard normal distribution, as P(z≥-1.95)=P(z≤1.95).

P(z\geq -1.95)=0.974

(g) P(−1.20 ≤ Z ≤ 2.00)

This is the same case as point a.

P(-1.20 \leq z \leq 2.00)= P(z

(h) P(1.01 ≤ Z ≤ 2.50)

This is the same case as point a.

P(1.01 \leq z \leq 2.50)= P(z

(i) P(1.20 ≤ Z)

This is the same case as point f.

P(z\geq 1.20)=0.115

(j) P(|Z| ≤ 2.50)

In this case, the z is expressed in absolute value. If z is positive, it has to be under 2.5. If z is negative, it means it has to be over -2.5. So this probability is translated to P|Z| < 2.50)=P(-2.5<z<2.5) and then solved from there like in point a.

P(|z|

7 0
3 years ago
Read 2 more answers
HELP ME QUICK PLEASEEEE solve for x: 2/x-2+7/x^2-4=5/x
Anna35 [415]

 

\displaystyle\\\\\frac{2}{x-2}+\frac{7}{x^2-4}=\frac{5}{x}\\\\\frac{^{x+2)}2~~~~~}{x-2}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2(x+2)}{(x-2)(x+2)}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2x+4+7}{x^2-4}=\frac{5}{x}\\\\\frac{2x+11}{x^2-4}=\frac{5}{x}\\\\5(x^2-4)=x(2x+11)\\\\5x^2-20=2x^2+11x\\\\5x^2-2x^2 -11x-20=0


\displaystyle\\3x^2-11x-20=0\\\\x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{11\pm\sqrt{121+4\cdot3\cdot20}}{2\cdot3}=\\\\=\frac{11\pm\sqrt{121+240}}{6}=\frac{11\pm\sqrt{361 }}{6}=\frac{11\pm19}{6}\\\\x_1=\frac{11-19}{6}=\frac{-8}{6}\\\\\boxed{\bf x_1=-\frac{4}{3}}\\\\x_2=\frac{11+19}{6}=\frac{30}{6}\\\\\boxed{\bf x_2=5}




3 0
3 years ago
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