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pychu [463]
3 years ago
12

What is the probability of randomly selecting either a club or a non-face card from a standard deck of cards?

Mathematics
1 answer:
aalyn [17]3 years ago
5 0
There are 13 clubs in a deck of card and 13 non face cards.
P(club of non face card) = 13/52 + 13/52 = 26/52 = 1/2
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Substitution means making one formula equal to a variable in the second formula to plug it in, and elimination means the cancelling out of one variable in order to make it easy to solve.

1. x-4(-x+5)=10            

  x+4x-20=10

  5x=30

  x=6, plug in 6 into original equation, y=-1

2. <u>-1(2x</u>-5y=6)

    <u>2x</u>+3y=-2    underlined means cancelled, so you add the rest up on each side

    5y+3y=-6+-2

    8y=-8

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   Infinite solutions

4. y=<u>4x</u>-3

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6 0
3 years ago
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Need help what’s the answer ASAP
Lesechka [4]

Answer:

V = 6.26 (2.5)

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Three different accounts are described below. Order the accounts according to their values after 10 years, from greatest to leas
faust18 [17]

Answer:

`1. You deposit $950 in an account that earns 9% annual interest compounded daily.

2.You deposit $1000 in an account that earns 8% annual interest compounded daily.

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Step-by-step explanation:

Trick question, number 3 says semiannually so it is only twice a year, so it is the least. In math papa or any algebra calculator, put in the two equations and compare which is greater, in this case, y = 950(1.09)^10 is greater than y=1000(1.08)^10

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amm1812
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4 0
3 years ago
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A. Use the​ one-mean t-interval procedure with the sample​ mean, sample​ size, sample standard​ deviation, and confidence level
PtichkaEL [24]

Answer:

a) (17.227, 22.773)

b) 2.773

c) 2.773

Step-by-step explanation:

Given:

Sample size, n = 25

Standard deviation, s = 4

Sample mean, x' = 20

Level of significance, a = 0.98 = 1 - 0.98 = 0.02

The degrees of freedom, df, for a t-distribution = n - 1 = 25 - 1 = 24

Using the t table, the Critical value = t_\alpha _/_2, _d_f = t_0_._0_2_/_2, _2_4 = t_0_._0_1_, _2_4 = 3.4668

Margin of error, E = t_\alpha _/_2, _d_f * \frac{\sigma}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773

Limits of 98% confidence interval, we have:

Lower limit : x' - M.E = 20 - 2.773 = 17.227

Upper limit: x' + M.E = 20 + 2.773 = 22.773

Therefore, (17.227, 22.773) is 98% confidence interval.

b) Let's the margin of error by taking half the length of the confidence interval.

Since we are to use half the length of CI, we have:

M.E = \frac{22.773 - 17.227}{2} = 2.773

c)M.E = t_\alpha _/_2, _d_f *s/\sqrt{n}

= t * \frac{s}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773

4 0
3 years ago
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