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3 years ago

7

How to write 2350 million in standard form ?

1 answer:

Arisa [49]3 years ago

6 0

Answer:

200,000,000 + 30,000,000 + 5,00,000

Step-by-step explanation:

you dont need to add the other numbers when you do that and it should be right

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This is because $10 is a greater percentage of $60 than it is of $90 ( if we calculate it is 16.66% in the first case and 11.11% in the second case).

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3 years ago

How to solve this problem

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3 years ago

Emily is entering a bicycle race for charity. her mother pledges $0.80 for every 0.5 mile she bikes. if emily bikes 14 miles, ho

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4 years ago

Read 2 more answers

between noon and midnight the temperature dropped 20 degrees Fahrenheit if the temperature was -8 degrees Fahrenheit at midnight

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3 years ago

Bob and Anna are planning to meet for lunch at Sally's Restaurant, but they forgot to schedule a time. Bob and Anna are each goi

Viktor [21]

Answer:

The expected cost is $8.75

<em></em>

Step-by-step explanation:

Given

$Time = \{1pm, 2pm, 3pm, 4pm\}$

$C_1 = \$5$ --- If Bob and Anna meet

$C_2 = \$10$ --- If Bob and Anna do not meet

Required

The expected cost of Bob's meal

First, we list out all possible time both Bob and Anna can select

We have:

$(Bob,Anna) = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3)$ $,(3,4),(4,1),(4,2),(4,3),(4,4)\}$

$n(Bob, Anna) = 16$

The outcome of them meeting at the same time is:

$Same\ Time = \{(1,1),(2,2),(3,3),(4,4)\}$

$n(Same\ Time) = 4$

The probability of them meeting at the same time is:

$Pr(Same\ Time) = \frac{n(Same\ Time)}{n(Bob,Anna)}$

$Pr(Same\ Time) = \frac{4}{16}$

$Pr(Same\ Time) = \frac{1}{4}$

The outcome of them not meeting:

$Different = \(Bob,Anna) = \{(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2)$

$,(3,4),(4,1),(4,2),(4,3)\}$

$n(Different) = 12$

The probability of them meeting at the same time is:

$Pr(Different) = \frac{n(Different)}{n(Bob,Anna)}$

$Pr(Different) = \frac{12}{16}$

$Pr(Different) = \frac{3}{4}$

The expected cost is then calculated as:

$Expected = C_1 * P(Same) + C_2 * P(Different)$

$Expected = \$5 * \frac{1}{4} + \$10 * \frac{3}{4}$

$Expected = \frac{\$5}{4} + \frac{\$30}{4}$

Take LCM

$Expected = \frac{\$5+\$30}{4}$

$Expected = \frac{\$35}{4}$

$Expected = \$8.75$

<em>The expected cost is $8.75</em>

4 0

3 years ago