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3 years ago

In the given circle the radius is 9 cm what is its diameter?

Mathematics

1 answer:

-Dominant- [34]3 years ago

5 0

Answer:

Step-by-step explanation:

The diameter is equal to twice the length of the radius

So if the radius is 9 then the diameter is 9 * 2 = 18

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Select three ratios that are equivalent to 5:2<br> Choose 3 answers:

Scorpion4ik [409]

Answer:

10:4

35:14

55:22

8 0

4 years ago

Compare the average score of a student at the beginning of a month and at the end of that month, if at first, it increased by 40

Leno4ka [110]

Answer:

122.5

Step-by-step explanation:

Let the student normal score be 100

In the beginning score increased by 40%

Hence score at the beginning =100+40 = 140%

Now before end score decreased by 25%

Decrease = 25% of 140 = 35

Hence score at the end of month = 140-35 = 105

Average score for the month

= Total of scores of end and beginning month /2

= $\frac{140+105}{2} \\=122.5$

4 0

3 years ago

Read 2 more answers

Square HIJK has a side length of 8 centimeters. Find the area of the square. Type a numerical answer in the space

tiny-mole [99]

The area of a square with side s is

Area = s²

Here

Area = 8² = 64 sq cm

Answer: 64

5 0

3 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

labwork [276]

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: $({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)\;=\;[({}^{4}C_1)^5]$