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Papessa [141]
3 years ago
5

A fair die is tossed once, what is the probability of obtaining neither 5 nor 2?​

Mathematics
1 answer:
Leya [2.2K]3 years ago
7 0

Answer:

4/6 or 66.666...%

Step-by-step explanation:

If you want to find the probability of obtaining neither a 5 nor a 2 you find how many times they occur and add them together in this case 5 occurs once and 2 also occurs once out of 6 numbers so 1/6 + 1/6 equals 2/6, you now know that 4/6 of them won't be a 5 nor a 2 and because it is a fair die the likelihood of it falling on a number is the same for all sides so the answer is 4/6 or 66.67%.

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liubo4ka [24]
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6 0
3 years ago
Read 2 more answers
CALCULUS EXPERT WANTED
Vlad [161]

a. Use the mean value theorem. 16 falls between 12 and 20, so

v'(16)\approx\dfrac{v(20)-v(12)}{20-12}=\dfrac{240-200}8=5

(Don't forget your units - 5 m/min^2)

b. v(t) gives the Johanna's velocity at time t. The magnitude of her velocity, or speed, is |v(t)|. Integrating this would tell us the total distance she has traveled whilst jogging.

The Riemann sum approximates the integral as

\displaystyle\int_0^{40}|v(t)|\,\mathrm dt=12\cdot200+8\cdot240+4\cdot220+16\cdot150=7600

If you're not sure how this is derived: we're given 5 sample points, so we can cut the interval [0, 40] into 4 subintervals. The lengths of each subinterval are 12, 8, 4, and 16 (the distances between each sample point), and the height of the rectangle approximating the area under the plot of |v(t)| is determined by the value of v(t) at each sample point, 200, 240, |-220| = 220, and 150.

c. Bob's velocity is given by B(t), so his acceleration is given by B'(t). We have

B'(t)=3t^2-12t

and at t=5 his acceleration is B'(5)=15 m/min^2.

d. Bob's average velocity over [0, 10] is given by the difference quotient,

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6 0
3 years ago
Find 8 and one third percent of 144
madam [21]
\bf 8\frac{1}{3}\implies \cfrac{8\cdot 3+1}{3}\implies \cfrac{25}{3}
\\\\\\
now\qquad \cfrac{\frac{25}{3}}{100}\implies \cfrac{\frac{25}{3}}{\frac{100}{1}}\implies \cfrac{25}{3}\cdot \cfrac{1}{100}\implies \cfrac{1}{3\cdot 4}\implies \cfrac{1}{12}
\\\\\\
8\frac{1}{3}\%\ of\ 144\implies  \left( \cfrac{\frac{25}{3}}{100} \right)\cdot 144\implies \cfrac{1}{12}\cdot 144\implies \cfrac{144}{12}\implies 12
3 0
3 years ago
Add.
lidiya [134]
The answer is c:
87.872
5 0
4 years ago
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