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irga5000 [103]
3 years ago
6

($1.75+2times $0.25 +5times $0.05) times 24

Mathematics
2 answers:
Anastaziya [24]3 years ago
8 0
(1.75 + 2 x 0.25 + 5 x 0.05) x 24= 17
KengaRu [80]3 years ago
4 0

Answer:

60

Step-by-step explanation:1.75+(2x0.25=.50)=2.25+(5x0.05=0.25)=2.50x24=60


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Use a known maclaurin series to obtain the maclaurin series for the given function. f(x = 8x2 tan?1(7x3
11Alexandr11 [23.1K]
Going out on a limb here and guessing that the function is

f(x)=8x^2\tan^{-1}(7x^3)

Please correct me if this isn't the case.

Recall that

\tan^{-1}x=\displaystyle\sum_{n\ge0}\frac{(-1)^nx^{2n+1}}{2n+1}

which converges for |x|.

It follows that

8x^2\tan^{-1}(7x^3)=8x^2\displaystyle\sum_{n\ge0}\frac{(-1)^n(7x^3)^{2n+1}}{2n+1}
=\displaystyle\sum_{n\ge0}\frac{56(-49)^nx^{6n+3}}{2n+1}
6 0
3 years ago
15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0
3 years ago
Which numbers are equivalent to 3/10
Sedbober [7]

Answer:

Multiply numerator and denominator with same number to gwt the answers . There are infinitely many numbers

3/10 × 2/2 = 6/20

3/10 × 5/5 = 15/50

3/10 × 10/10 = 30/100

And so on

8 0
3 years ago
Two step equation<br><br> n - 5/2 =5
Verdich [7]

Answer:

Step-by-step explanation:

two step equation

n - 5/2 = 5

n = 5 + 5/2

n = 5 + 2.5

n = 7.5

-----------

check

7.5 - 5/2 = 5

7.5 - 2.5 = 5

5 = 5

the answer is good

3 0
1 year ago
Skylar makes turkey sandwiches for a family reunion. He purchases 5 pounds 4 ounces of sliced turkey for $26.88. What is the uni
Advocard [28]

Answer:

The unit cost per ounce for turkey skylar is $ 0.316  

Step-by-step explanation:

The total amount of sliced turkey = 5 pounds + 4 ounces

The total cost of turkey slice = $26.88

Now, as we know

1 pounds = 16 ounces

∴ 5 pounds = 16 × 5 = 80 ounces

So, the total amount is 5 pounds + 80 pounds = 85 pounds

∵ 85 pounds of turkey sliced cost $ 26.88

∴ 1 pounds of turkey sliced cost $ \frac{26.88}{85}  = $ 0.316

Hence the unit cost per ounce for turkey skylar is $ 0.316  Answer

8 0
3 years ago
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