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3 years ago

How much lower is -127 than 1578

Mathematics

1 answer:

natta225 [31]3 years ago

3 0

The answer is 1705 hope this helped

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The repeating decimal,0.27 repeting, is converted to the fraction 3/x. What is the value of x in the fraction?

Nuetrik [128]

Answer:5/18=.2777777777777

Step-by-step explanation:

.27777 (7 repeating)

100x=27.7777

10x= 2.7777

Subtract

90x=25

x=25/90=5/18

Check

5/18=.2777777777777

Hope that helps.

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4 years ago

Someone please help me! ‘The value of X is...’

8_murik_8 [283]

This is called a double equation. If you solve for X, you would get 'any whole number.'

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4 years ago

A(n) _______ is part of a mathematical expression consisting of a constant, or a variable, or the product of a number and a vari

Fofino [41]

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3 years ago

Find the minimum value of<br> C = 6x + 7y

statuscvo [17]

The minimum value of c is 46

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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

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