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4 years ago

A map scales reads: 2 inches = miles. Which distance does 6 inches represent

Mathematics

1 answer:

djverab [1.8K]4 years ago

3 0

3 miles
2 inches per mile and there is 6 inches
6 divided by 2 is 3

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Erica is making feathered caps for her school play. Each cap must have 3 feathers. Write an equation that represents the number

FrozenT [24]

Answer:

$f=3c$

Step-by-step explanation:

Given:

Number of feathers = $f$

Number of caps = $c$

Number of feathers in each cap = 3

Now, solving by setting up a proportion.

1 cap has 3 feathers. 'c' caps has 'f' feathers.

This gives,

$\dfrac{1}{3}=\dfrac{c}{f}$

Solving by cross multiplication, we get:

$1\times f=3\times c\\\\f=3c$

Therefore, the equation representing the number of feathers 'f' Erica needs to make 'c' caps is given as:

$f=3c$

8 0

3 years ago

Harriet is helping her brother paint an art project like the one shown below. They plan to paint only the lateral faces. One qua

Reil [10]

I like cheese and cheese is good

8 0

3 years ago

Use a positive or negative number to represent a stock market loss of $1,500.

Natasha_Volkova [10]

Answer:

negative

Step-by-step explanation:

negative because you are losing something

7 0

3 years ago

We need a minimum of 30 book sales by today. If 10 books were sold yesterday, what are the possible numbers for sales today

amm1812

I mean it would be A and D because I don't know why they are both the same answer. If 10 books were sold yesterday, at LEAST 20 are needed to make the goal.

6 0

3 years ago

A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one

TiliK225 [7]

In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are $\binom44\binom{48}1=48$ possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by $\binom{13}1=13$ . So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

$\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024$

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing $\binom41\binom{48}4=778,320$ possible hands. Exactly 2 aces are drawn in $\binom42\binom{48}3=103,776$ hands. And so on. This gives a total of

$\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656$

possible hands containing at least 1 ace, and hence B occurs with probability

$\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412$

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. $P(A\cap B)=P(A)P(B)$ . This happens if

the hand has 4 aces and 1 non-ace, or
the hand has a non-ace 4-of-a-kind and 1 ace

The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So $A\cap B$ consists of 96 possible hands, which occurs with probability

$\dfrac{96}{\binom{52}5}\approx0.0000369$

and so the events A and B are NOT independent.

4 0

3 years ago