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3 years ago

Pls help me asap pls

Mathematics

2 answers:

julia-pushkina [17]3 years ago

8 0

Answer:

Im pretty sure it's 33 units

Step-by-step explanation:

sorry if im wrong mark me brainliest if im right

hoa [83]3 years ago

8 0

Answer:

I also believe the answer is 33 units squared

Step-by-step explanation:

If you count all the whole boxes you get 30. then two halved make a whole so if there are six half squares then there are 3 whole squares. 30+3=33. always remember the units at the end. I hope Im right sorry if wrong or not helpful

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Justin bought his truck new for $32,000. Its value decreases 9.0% each year. Choose the letter for the best answer.

Pavel [41]

Answer:

Justin bought his truck new for $32,000. Its value decreases 9.0% each year. Choose the letter for the best answer.

5. Which function represents the yearly value of Justin's truck?

A f(t) = 32,000(1 + 0.9)^t

B f(t) = 32,000(1 - 0.9)^t

C f(t) = 32,000(1 + 0.09)^t

D f(t) = 32,000(1 - 0.09)^t 

6. When will the value of Justin's truck fall below half of what he paid for it?

F In 6 years

G In 8 years 

H In 10 years

J In 12 years

Step-by-step explanation:

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3 years ago

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Zepler [3.9K]

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4 years ago

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3 years ago

Which equals the product of (x-3)(2x + 1)? 222 – 7x - 3 22 – 5x - 3 3x = 2 6,2

Katena32 [7]

Answer:

Step-by-step explanation:

$2 {x}^{2} - 5x - 3$

4 0

3 years ago

In a shipment of 20 packages, 7 packages are damaged. The packages are randomly inspected, one at a time, without replacement, u

horsena [70]

Answer:

$\large\boxed{\large\boxed{0.119}}$

Explanation:

You need to find the probability that exactly three of the first 11 inspected packages are damaged and the fourth is damaged too.

1. Start with the first 11 inspected packages:

a) The number of combinations in which 11 packages can be taken from the 20 available packages is given by the combinatory formula:

$C(m,n)=\dfrac{m!}{m!(m-n)!}$

$C(20,11)=\dfrac{20!}{11!\cdot(20-11)!}$

b) The number of combinations in which 3 damaged packages can be chossen from 7 damaged packages is:

$C(7,3)=\dfrac{7!}{3!\cdot(7-3)!}$

c) The number of cominations in which 8 good packages can be choosen from 13 good pacakes is:

$C(13,8)=\dfrac{13!}{8!\cdot(13-8)!}$

d) The number of cominations in which 3 damaged packages and 8 good packages are chosen in the first 11 selections is:

$C(7,3)\times C(13,8)$

e) The probability is the number of favorable outcomes divided by the number of possible outcomes, then that is:

$\dfrac{C(7,3)\times C(13,8)}{C(20,11)}$

Subsituting:

$\dfrac{\dfrac{7!}{3!\cdot(7-3)!}\times \dfrac{13!}{8!\cdot(13-8)!}}{\dfrac{20!}{11!\cdot(20-11)!}}$

$=\dfrac{\dfrac{7!}{3!\cdot 4!}\times \dfrac{13!}{8!\cdot 5!}}{\dfrac{20!}{11!\cdot 9!}}=0.26818885$

2. The 12th package

The probability 12th package is damaged too is 7 - 3 = 4, out of 20 - 11 = 9:

3. Finally

The probability that exactly 12 packages are inspected to find exactly 4 damaged packages is the product of the two calculated probabilities:

$0.26818885\times 4/9=0.119$

6 0

3 years ago