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3 years ago

Casey has a litter of 5 puppies and 3 of them are gray. At this rate, how many puppies would be gray in a litter of 30 puppies?

Mathematics

1 answer:

deff fn [24]3 years ago

6 0

Answer:

18 gray puppies

Step-by-step explanation:

Think of this as finding equivalent fractions/ratios. 3/5 puppies are gray. what over 30 puppies would be gray? to find 3/5=?/30, you multiply the numerator and denominator by 6. 3 times 6 is 18.

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What type of correlation is shown in this scatterplot?

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You have 100 people. 90 have French heritage, 80 have English heritage and 75 have Native American heritage, how many have herit

USPshnik [31]

My answer is 82. I added up 90 + 80 + 75 = 245. I then divided that by 3 because there are 3 different cultures. And the number was 81. something. So, I rounded and got 82. But I'm not sure if that's the right answer.

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4 years ago

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A. what is asked?

Andru [333]

Answer:

A. How much Anna weights in kilograms is what is asked (Btw is it her or his?)

B. The values given is 54000 (most likely in grams )

C. The operation that is going to be used is 1000 since kilograms is 1000 grams

D. The solution is 54 kilograms

E. To get 54 kilograms to grams we can just multiply by 1000 again which will result into 54000 thus the answer is correct

Step-by-step explanation:

The question is asked when there is a "what" or a problem statement

The only value or digits given is 54000

in the Metric Scale Kilograms is 1000 times larger in value than grams thus 1000

Divide 54000 with 1000 which is 54

Um.... just multiply 1000 ._. (That's the only way I can check it)

Hope this helps!

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3 years ago

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A manufacturing company has two retail outlets. It is known that 30% of all potential customers buy

aleksandrvk [35]

Answer:

(a) A y P(A) = 0.4 (b) $\bar{B}$ y $P(\bar{B})$ =0.5 (c) $\bar{A}$ ∪ $\bar{B}$ y P( $\bar{A}$ ∪ $\bar{B}$ ) = 0.9 (d) $\bar{A}$ ∩ $\bar{B}$ y P( $\bar{A}$ ∩ $\bar{B}$ )=0.2

Step-by-step explanation:

A was defined as the event that a potential customer, randomly chosen, buys from outlet 1 in the original problem statement. We know that B denotes the event that a randomly chosen customer buys from outlet 2. So

P( $A\cap \bar{B}$ ) = 0.3, P( $B\cap \bar{A}$ ) = 0.4 and P( $A\cap B$ ) = 0.1

(a) P(A) = P( $A\cap (B\cup\bar{B})$ ) = P( $A\cap B$ ) + P( $A\cap \bar{B}$ ) = 0.1 + 0.3 = 0.4

(b) P(B) = P( $B\cap (A\cup\bar{A})$ ) = P( $B\cap A$ ) + P( $B\cap \bar{A}$ ) = 0.1 + 0.4 = 0.5

P( $\bar{B}$ ) = 1-P(B) = 1-0.5 = 0.5

(c) The customer does not buy from outlet 1 is the complement of A, i.e., $\bar{A}$ , and the customer does not buy from outlet 2 is the complement of B, i.e., $\bar{B}$ , so, the customer does not buy from outlet 1 or does not buy from outlet 2 is $\bar{A}$ ∪ $\bar{B}$ and P( $\bar{A}$ ∪ $\bar{B}$ ) = P( $(A\cap B)^{c}$ ) by De Morgan's laws

P( $(A\cap B)^{c}$ ) = 1-P(A∩B)=1-0.1=0.9

(d) The customer does not buy from outlet 1 is the complement of A, and the customer does not buy from outlet 2 is the complement of B, so we have that the statement in (d) is equivalent to $\bar{A}$ ∩ $\bar{B}$ and P( $\bar{A}$ ∩ $\bar{B}$ ) = P( $(AUB)^{c}$ ) by De Morgan's laws, and

P( $(AUB)^{c}$ ) = 1-P(A∪B)=1-[P(A)+P(B)-P(A∩B)]=1-[0.4+0.5-0.1]=1-0.8=0.2

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3 years ago