How does a triangular prism look

Mathematics

1 answer:

irina1246 [14]3 years ago

6 0

Answer:

its like a 3d rectangle triangle

Step-by-step explanation:

yes

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6lbs of apples for 33.60

Mkey [24]

Answer:

If that's the whole question then what exactly are we trying to solve here?

Step-by-step explanation:

5 0

3 years ago

Your brother is 1/5 your age. Your sister is 9 years older than your brother. Your sister is 11 years old. Write and solve an eq

Papessa [141]

Brother=b
b+9=11
b=2
2=1/5a
10=a
a=10

The answer is 10

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3 years ago

Pauley graphs the change in temperature of a glass of hot tea over time. he sees that the function appears to decrease quickly a

babunello [35]

Non-linear graph best describes this function.

<h3>Why nonlinear?</h3>

As described in the question, Pauley graphs the change in temperature of a glass of hot tea over time. He sees that the function appears to decrease quickly at first, then decrease more slowly as time passes. It is linear because the graph decreases over time. It is linear because there is both an independent and a dependent variable but it doesn't satisfies one condition of being linear. Thus, It's Nonlinear because linear functions are increasing functions. It is nonlinear because linear functions increase or decrease at the same rate.

To learn more about nonlinear function visit:

brainly.com/question/762799

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5 0

2 years ago

You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob

Ilia_Sergeevich [38]

Answer:

a) No

b) 42%

c) 8%

d) X 0 1 2

P(X) 42% 50% 8%

e) 0.62

Step-by-step explanation:

a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.

b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6

P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7

P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

c) P(win first game) = 0.4

P(win second game) = 0.2

P(win both games) = P(win first game) × P(win second game) = 0.4 × 0.2 = 0.08 = 8%

d) X 0 1 2

P(X) 42% 50% 8%

P(X = 0) = P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

P(X = 1) = [ P(lose first game) × P(win second game)] + [ P(win first game) × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%

e) The expected value $\mu=\Sigma}xP(x)= (0*0.42)+(1*0.5)+(2*0.08)=0.66$