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2 years ago

7

If 18 children share 452 candies equally, how many candies will be left unshared?

2 answers:

NeX [460]2 years ago

5 0

Answer:

452 divided by 18 equals

25 with a remainder of 2

Step-by-step explanation:

kirza4 [7]2 years ago

4 0

Answer:

Around 25

Step-by-step explanation:

452 candies/18 children

= 25.11111111

Meaning there are around 25 left

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Which statement is true about the labeled function? F(x) < 0 over the Interval (-infinity, -4)

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Answer:

4

Step-by-step explanation:

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3 years ago

Y = x^3 + 5x^2 + 3x x= -3

dexar [7]

Answer:

9

Step-by-step explanation:

-3^3 +5*(3)^2 +3*-3

-27+5*9+(-9)

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3 years ago

Read 2 more answers

Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen

SSSSS [86.1K]

Answer:

a.) 0.5

b.) 0.66

c.) 0.83

Step-by-step explanation:

As given,

Total Number of Batteries in the drawer = 10

Total Number of defective Batteries in the drawer = 4

⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6

Now,

As, a sample of 3 is taken at random without replacement.

a.)

Getting exactly one defective battery means -

1 - from defective battery

2 - from non-defective battery

So,

Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ = $\frac{4!}{1! (4 - 1 )!}$ × $\frac{6!}{2! (6 - 2 )!}$

= $\frac{4!}{(3)!}$ × $\frac{6!}{2! (4)!}$

= $\frac{4.3!}{(3)!}$ × $\frac{6.5.4!}{2! (4)!}$

= $4$ × $\frac{6.5}{2.1! }$

= $4$ × $15$ = 60

Total Number of possibility = ¹⁰C₃ = $\frac{10!}{3! (10-3)!}$

= $\frac{10!}{3! (7)!}$

= $\frac{10.9.8.7!}{3! (7)!}$

= $\frac{10.9.8}{3.2.1!}$

= 120

So, probability = $\frac{60}{120} = \frac{1}{2} = 0.5$

b.)

at most one defective battery :

⇒either the defective battery is 1 or 0

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 0 , then 3 non defective

Possibility = ⁴C₀ × ⁶C₃

= $\frac{4!}{0! (4 - 0)!}$ × $\frac{6!}{3! (6 - 3)!}$

= $\frac{4!}{(4)!}$ × $\frac{6!}{3! (3)!}$

= 1 × $\frac{6.5.4.3!}{3.2.1! (3)!}$

= 1× $\frac{6.5.4}{3.2.1! }$

= 1 × 20 = 20

getting at most 1 defective battery = 60 + 20 = 80

Probability = $\frac{80}{120} = \frac{8}{12} = 0.66$

c.)

at least one defective battery :

⇒either the defective battery is 1 or 2 or 3

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 2 , then 1 non defective

Possibility = ⁴C₂ × ⁶C₁

= $\frac{4!}{2! (4 - 2)!}$ × $\frac{6!}{1! (6 - 1)!}$

= $\frac{4!}{2! (2)!}$ × $\frac{6!}{1! (5)!}$

= $\frac{4.3.2!}{2! (2)!}$ × $\frac{6.5!}{1! (5)!}$

= $\frac{4.3}{2.1!}$ × $\frac{6}{1}$

= 6 × 6 = 36

If the defective battery is 3 , then 0 non defective

Possibility = ⁴C₃ × ⁶C₀

= $\frac{4!}{3! (4 - 3)!}$ × $\frac{6!}{0! (6 - 0)!}$

= $\frac{4!}{3! (1)!}$ × $\frac{6!}{(6)!}$

= $\frac{4.3!}{3!}$ × 1

= 4×1 = 4

getting at most 1 defective battery = 60 + 36 + 4 = 100

Probability = $\frac{100}{120} = \frac{10}{12} = 0.83$

3 0

3 years ago

A. Sales are (increasing/decreasing)…. _____….(purchases, purchases/month, months/purchases, $/purchases, purchases/$, $, months

lord [1]

Answer:

Sales are (increasing/decreasing)…. _____….(purchases, purchases/month, months/purchases, $/purchases, purchases/$, $, months)

D. (January, February, March, April, May, June, July, August, September, October, November, December)

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2 years ago

You recently invested $400,000 of your savings in a security issued by a large company. The security agreement pays you 12% per

ANTONII [103]

Answer:

$722444.49386776

Step-by-step explanation:

Use Compound Intrest Formula

$a = p(1 + \frac{r}{n} ) {}^{nt}$

where p is the original amount.
R is the amount of percentage compounded
N is amount of times compounded per year.
T is how long the interest last.

P is 400,00p

T is 12% or 0.12

N is 4 since it is compounded quarterly

T is 5.

Plug the values in

$400000(1 + \frac{0.12}{4} ) {}^{20}$

Ypu get

$722444.49386776

6 0

3 years ago