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3 years ago

14

in a talent competition, half of the contestants are eliminated in each round. At the end of the nth round, 32 contestants remai

n. If there were 1,024 contestants at the start of the competition, what is the value of n?

1 answer:

Murrr4er [49]3 years ago

3 0

Here, first tern is 1024, common ratio = 1/2 and last term = 32

32 = 1024((1/2)^n
(1/2)^n = 32/1024 = 1/32
(1/2)^n = (1/2)^5

Therefore, n = 5

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How do we refer to self-contained state actions: monadic. dyadic. triadic. k-adic?

Assoli18 [71]

This state action is referred to as monadic. This is a function or a relation with an arity of one. A monad can relate an algebraic theory into a <span>composition of a function though its power is not always apparent.</span>

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3 years ago

A french class has a total of 25 student.The number of males is 7 more than the number of females how many males and how many fe

Lady_Fox [76]

Answer:

16 males and 9 females

Step-by-step explanation:

To solve this we can use a system of equations.

Let's start by naming the number of females x.

The number of males would then be y.

<u>Using these variables, we can set up 2 equations using info provided:</u>

A french class has a total of 25 students, -> x+y=25

The number of males is 7 more than the number of females -> x+7=y

Use substitution to solve.

<u>From the second equation:</u>

x+7=y

Subtract 7 from both sides.

x=y-7

Substitute that into the first equation.

x+y=25

y-7+y=25

Combine like terms.

2y-7=25

Add 7 to both sides.

2y=32

Divide both sides by 2.

y=16

Substitute y=16 into equation 2.

x+7=y

x+7=16

Subtract 7 from both sides.

x=9

Therefore, there are 16 males and 9 females in the french class.

3 0

2 years ago

Read 2 more answers

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65

erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 2.65

$\sigma$ = standard deviation = 0.85

n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

P(X bar <= 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z < 2.06) = 0.98030

P(X bar <= 2.65) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3) = 0.98030 - 0.5 = 0.4803 .

8 0

3 years ago

Can you solve this for me please

Hatshy [7]

Answer:

Graph 1

Step-by-step explanation:

f(x) = __

In this equation (f(2) = 4) the 2 is the x value. All this question is asking is which graph shows a y value of 4 when x is 2.

4 0

3 years ago

Ian has 12.2 ounce bottle of ketchip. He uses 1/20 of the ketchup every time he has a buffalo burger How many ounces of ketchup

tatiyna

If Ian uses 1/20 of the ketchup 4 times, this is equal to 4/20, or $\frac{4}{20}$ .

$\frac{4}{20}$ can be simplified to $\frac{1}{5}$ , which means we need to work out $\frac{1}{5}$ of 12.2

$\frac{1}{5}$ is equivalent to 20%, which as a decimal is 0.2

0.2 x 12.2 = 2.44 - however, this is how much he has used, and we need to figure out how much is left.

12.2 - 2.44 = 9.76

So, Ian has 9.76 oz remaining :)

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3 years ago