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3 years ago

5)

Mathematics

2 answers:

Evgen [1.6K]3 years ago

5 0

Answer:

C) 8 miles for hour

Step-by-step explanation:

1/4 mile in 1/32 hr

unit rate would be x miles in 1 hour.

A complex fraction is a fraction that has a fraction in the numerator OR denominator OR both.

(1/4) / (1/32) =(1/4) * (32/1) = 8 miles in 1 hour

EleoNora [17]3 years ago

3 0

9514 1404 393

Answer:

C) 8 miles per hour

Step-by-step explanation:

To find miles per hour, divide miles by hours.

(1/4) ÷ (1/32) = 1/4 × 32/1 = 32/4 = 8 . . . . miles per hour

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What is the probality of rolling a sum of 5 if two regular 6 -sides number cubes are rolled

Dmitry [639]

Each die has six numbers, the total number of possible combinations is 6 * 6 = 36

The combinations in which you can get a sum of 5 are: 1 &4, 2 & 3, 4 &1, 3 &2

There are 4 different combinations of getting a sum of 5.

The probability would be 4/36, which reduces to 1/9

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4 years ago

general admission to the fair is 10.00. each ride costs 4.75. if lisa has 75.00 to spend at the fair, what method can be used to

Elis [28]

Answer:

well the answer is 13

Step-by-step explanation:

The way you can figure this out is subtract 10 from 75 to get 65 and then you divide 65 by 4.75 to get 13 with a huge decimal but you just have 13 because you cant go for part of a ride.

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3 years ago

-1/6b=3 b= A. -18 B. -2 C 2 D 18

goldenfox [79]

$-\frac{1}{6}b = 3\\\\-\frac{1}{6}b\times -6 = 3\times -6\\\\ b = -18$

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4 years ago

Five cards are dealt from a standard 52-card deck. (a) What is the probability that we draw 1 ace, 1 two, 1 three, 1 four, and 1

Rudik [331]

Answer:

Step-by-step explanation:

As there are total 52 cards in a deck and we have to draw a set of 5 cards, we can use the formula of combination to find the total number of possible ways of drawing 5 cards.

Number of ways to draw 5 cards = $N_T$

$N_T\;=\;({}^NC_k)\\\\N_T\;=\;({}^{52}C_5)\\\\N_T\;=\;2,598,960$

(a) Assuming the cards are drawn in order (would not affect the probability). The of getting Ace, 2, 3, 4 and 5 can be obtained by multiplying the probability of getting cards below 6 (20/52) with the probability of getting 5 different cards (4 choices for each card).

$P(a)\;=\;\frac{20}{52}*\frac{4}{52}*\frac{4}{51}*\frac{4}{50}*\frac{4}{49}*\frac{4}{48}\\\\P(a)\;=\; 1.3133*10^{-6}$

(b) For a straight we require our set to be in a sequence. The choices for lowest value card to produce a sequence are ace, 2, 3, 4, 5, 6, 7, 8, 9, or 10. Hence, the number of ways are $({}^{10}C_1)$ .

For each card we can draw from any of the 4 sets. It can be described mathematically as: $({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)\;=\;[({}^{4}C_1)^5]$

Therefore, the total outcomes for drawing straight are:

$N_S\;=\;({}^{4}C_1)*({}^{4}C_1)^5\;=\;10240$

Thus, the probability of getting a straight hand is:

$P(b)\;=\;\frac{N_S}{N_T}\\\\P(b)\;=\;\frac{10240}{2598960}\\\\P(b)\;=\; 0.0039$

6 0

3 years ago

To show that polygon ABCDE is congruent to polygon FGHIJ, a must be used to make the two polygons coincide. A sequence of two tr

Sati [7]

Hello.

The minimum number of rigid transformations required to show that polygon ABCDE is congruent to polygon FGHIJ is 2 (translation and rotation).

A rotation translation must be used to make the two polygons coincide.

A sequence of transformations of polygon ABCDE such that ABCDE does not coincide with polygon FGHIJ is a translation 2 units down and a 90° counterclockwise rotation about point D

Have a nice day

7 0

3 years ago

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