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3 years ago

-99,-66,-33... is an arithmetic sequence. What will the 11th term be?

Mathematics

1 answer:

m_a_m_a [10]3 years ago

5 0

Answer:

231

Step-by-step explanation:

-99, -66, -33, 0, 33, 66, 99, 132, 165, 198, 231\

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550000000000000000 • 0.00000000044

nasty-shy [4]

Answer:

550000000000000000 0.00000000044

Step-by-step explanation:

000000Oo

5 0

2 years ago

The expression 10a+6c gives the cost (in dollars) for a adults and c children to eat at a buffet restaurant. A. The cost for 1 a

Liono4ka [1.6K]

Answer: The total cost is $\$62$ assuming the cost for 1 adult is $\$5$ and the cost for 1 child is $\$2$

Step-by-step explanation:

Assuming the cost for 1 adult is $\$5$ and the cost for 1 child is $\$2$ :

$a=\$5$ and $c=\$2$

Then the expression that gives the total cost $10 a+6c$ is solved as:

$10(\$5)+6(\$2)=\$62$

4 0

3 years ago

Question 1 of 25

Murrr4er [49]

The company should expect 235 in sales

plug in 50 for x, 2.1 times 50 is 105, 105 + 130 is 235, y = 35.

7 0

2 years ago

urn I contains 1 red chip and 2 white chips; urn II contains 2 red chipsand 1 white chip. One chip is drawn at random from urn I

vitfil [10]

Answer:

Following are the answer to this question:

Step-by-step explanation:

In the question first calls the W if the transmitted chip was white so, the W' transmitted the chip is red or R if the red chip is picked by the urn II.

whenever a red chip is chosen from urn II, then the probability to transmitters the chip in white is:

$P(\frac{w}{R}) = \frac{P(W\cap R)}{P(R)} \ \ \ \ \ _{Where}\\\\P(R) = P(W\cap R) + P(W'\cap R) \\$

The probability that only the transmitted chip is white is therefore $P(W) = \frac{2}{3}\\$ , since urn, I comprise 3 chips and 2 chips are white.

But if the chip is white so, it is possible that urn II has 4 chips and 2 of them will be red since urn II and 2 are now visible, and it is possible to be: $P(\frac{R}{W}) = \frac{2}{3}$

$P(W\cap R) = P(W) \times P(\frac{R}{W}) \\$

$= \frac{2}{3}\times \frac{2}{4} \\\\= \frac{2}{3}\times \frac{1}{2} \\\\= \frac{2}{3}\times \frac{1}{1} \\\\=\frac{1}{3}\\\\= 0.333$

Likewise, the chip transmitted is presumably red $(P(W')= \frac{1}{3})$ and the chip transferred is a red chip of urn II $(P(\frac{R}{W'})= \frac{3}{4}$ , and a red chip is likely to be red $(\frac{R}{W'})$ .