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Arte-miy333 [17]
3 years ago
9

Simplify each expression. Write each answer in scientific notation. (2 * 10^3)(3 * 10^2)

Mathematics
1 answer:
spin [16.1K]3 years ago
3 0
The answer to that equation is 6*10^5
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Period of y=2tan (2x+90) -1
krek1111 [17]
20 is the answer
I hope this help
4 0
3 years ago
The price of an item has dropped to
Deffense [45]
60% decrease.

$105-$42 = $63
$63/$105 = 0.6
0.6 x 100 = 60
8 0
3 years ago
How do you solve this
yawa3891 [41]

Answer:

(a).\:\:\int _1^{-1}3f\left(x\right)dx\:=-12   is the final answer.

Step-by-step explanation:

See the attachment below.

Best Regards!

3 0
4 years ago
Read 2 more answers
Please I need help
GenaCL600 [577]

Answer:

Hi! You just have to take his salary as a unity (one) and substract the fractions you are given. For example; lets say you have a cake (one) and then you eat 1/8 of it. You would be left with  1-\frac{1}8} of the original cake.

Step-by-step explanation:

Salary:     1

Spent \frac{2}{5} :     1-\frac{2}{5}

Spent \frac{1}{4}:      1-\frac{2}{5} -\frac{1}{4}

Spent \frac{1}{8}:      1-\frac{2}{5} -\frac{1}{4} -\frac{1}{8}

REMAINING: 1-\frac{2}{5} -\frac{1}{4} -\frac{1}{8}

Operate using LCM:  \frac{40-16-10-5}{40} =  \frac{9}{40}

The answer is (d)

7 0
3 years ago
You are to play the following game of cards. Cards are worth their face value, Jacks, Queens and Kings are also worth 10 and Ace
ryzh [129]

Answer:

$82.875

Step-by-step explanation:

From the given information:

Assume F is used to denote the two cards;

If there are four aces among 52 playing cards, the chance of selecting the first ace is =  \dfrac{4}{52}

After selecting the first ace, we have only 3 aces remaining and a total of 51 playing cards. Thus, the chance of selecting the second ace will be \dfrac{3}{51}

By applying product rule, we can determine the chance of selecting two aces without replacement as follows:

i.e.

P(F=22)=\dfrac{4}{52}\times \dfrac{3}{51}

= \dfrac{1}{221}

The probability of getting one ace, one face is:

P(F=21) =( \dfrac{4}{52}\times \dfrac{12}{51})+( \dfrac{12}{52}\times \dfrac{4}{51})

P(F=21) = \dfrac{4}{221}\times\dfrac{4}{221}

P(F=21) = \dfrac{8}{221}

Since there are 4 aces, 4 nine, and 12 faces in a card deck

The probability of getting one ace, one nine, or two faces now will be:

P(F=20) = (\dfrac{4}{52} \times \dfrac{4}{51})+ (\dfrac{4}{52} \times \dfrac{4}{51}) + (\dfrac{12}{52} \times \dfrac{11}{51})

P(F=20) = (\dfrac{53}{663}  )

Now, the probability  of at least 20 now is:

\text{P(F at least 20)} = \dfrac{1}{221}+\dfrac{8}{221}+\dfrac{53}{663}

\text{P(F at least 20)} = \dfrac{80}{663}

If H represents the amount of prize of the expected winnings:

Then;

(H - 10) (\dfrac{80}{663}) + (-10)(\dfrac{663-80}{663}) = 0

\dfrac{80(H-10)}{663}-\dfrac{5830}{663}=0

\dfrac{80(H-10)}{663}=\dfrac{5830}{663}

80H - 800 = 5830

80H = 5830 +800

80H = 6630

H = 6630/80

H = $82.875

The prize should be $82.875 to make a winning positive.

4 0
3 years ago
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