What’s 36 x 99? I don’t know, do you?

Mathematics

2 answers:

user100 [1]3 years ago

8 0

Answer:

3564

Step-by-step explanation:

calculator, dude.

Lapatulllka [165]3 years ago

4 0

Answer:

Its 3,464 *smh

If you had 99 barrels of oil, then multiplied it why 36, how much would u have?

Answer - 3,464

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I REALLY NEED HELP! PLEASE AND THANK YOU. ITS ALGEBRA 2

umka2103 [35]

$f(x) = \dfrac{x+1}{x^2-7x-8} \\~\\~\\f(x) = \dfrac{x+1}{(x-8)(x+1)} \\~\\~\\f(x) = \dfrac{1}{x-8}$

$\text{avg} = \dfrac{f(a)-f(b)}{a-b} \\~\\~\\\text{avg} = \dfrac{\left(\dfrac{1}{3-8}\right)-\left(-\dfrac{1}{2+8}\right)}{3-(-2)} \\~\\~\\\text{avg} = \dfrac{\left(\dfrac{1}{-5}\right)-\left(-\dfrac{1}{10}\right)}{5} \\~\\~\\\text{avg} = \dfrac{1}{2}-1 \\~\\~\\ \text{avg} = -\dfrac{1}{2}$

3 0

3 years ago

Read 2 more answers

Suppose C1;C2;C3 are three didifferent biased coins, whose probability of heads equals 0.4, 0.5, and 0.2 respectively. Suppose c

allochka39001 [22]

Solution :

$P(A \cap B) = P(\text{coin}\ C_1 \text{ chosen and 4 heads in 10 trails})$

$= \frac{1}{3} \times ^{10}C_4 \times 0.4^4 \times 0.6^6$

= 0.0836

P(B)= P(coin $C_1$ chosen and 4 heads in 10 trails) + P(coin $C_2$ chosen and 4 heads in 10 trail) + P(coin $C_3$ chosen and 4 heads in 10 trail)

$=\frac{1}{3}\times ^{10}C_4 \times 0.4^4 \times 0.6^6 + \frac{1}{3}\times ^{10}C_4 \times 0.5^4 \times 0.5^6 + \frac{1}{3}\times ^{10}C_4 \times 0.2^4 \times 0.8^6$