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3 years ago

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Which is a word phrase for the variable expression? 3(n – 1) A. three times the quotient of a number n and 1 B. three more than

the difference of a number n and 1 C. one less than the product 3 and a number n D. three times the difference of a number n and 1

1 answer:

sergey [27]3 years ago

8 0

D. three times the difference of a number n and 1

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2 years ago

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Find the value of x. Round the length to nearest tenth.

Lady_Fox [76]

From the law of sines, we have

$l = h\sin(\theta) = h\cos(\alpha)$

where $l$ is the leg we're interested in, $h$ is the hypothenuse, $\theta$ is the angle opposite to $l$ , and $\alpha$ is the angle between $l$ and $h$ .

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3 years ago

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Marion bought 14 packages of plastic frog and lizard fishing lures. She bought frog lures in packages of 4 and lizard lures in p

tatuchka [14]

Answer:

Marion bought 6 frog lures and 8 lizard fishing lures.

Step-by-step explanation:

Let $f$ represent frog lures and $l$ represent lizard lures.

The total packages Marion bought is

$f+l=14...eqn1$

If she bought frog lures in packages of 4 and lizard lures in packages of 6 for a total of 72 lures, then

$4f+6l=72...eqn2$

We make $f$ the subject in equation 1 and put it into equation 2 to get;

$f=14-l...eqn3$

We put equation 3 into equation 2 to get;

$4(14-l)+6l=72$

$\Righttarrow 2(14-l)+3l=36$

We expand the bracket to get;

$28-2l+3l=36$

$-2l+3l=36-28$

$l=8$

We put $l=8$ into equation 3 to get;

$f=14-8$

$f=6$

Therefore Marion bought 6 frog lures and 8 lizard fishing lures.

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3 years ago

Who replaced circles with ellipses in a heliocentric model of the universe

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3 years ago

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro

Vinvika [58]

Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

0.95 probaiblity of hitting a target

This means that $p = 0.95$

10 targets

This means that $n = 10$

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

$P(X = 10) + P(X < 10) = 1$

We want P(X < 10). So

$P(X < 10) = 1 - P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987$

$P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401$

40.1% probability that he will miss at least one of them

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3 years ago