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3 years ago

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Lexie has dimes and ten dollar bills in her pocket. The number of dimes is six more than two times the number of ten dollar bill

s. Let t represent the number of ten dollar bills. Write an expression for the number of dimes

1 answer:

madam [21]3 years ago

3 0

Answer:

d=6+2t

Step-by-step explanation:

Let

t=number of ten dollar bills

d=number of dimes

Number of dimes is six more than two times the number of ten dollar bills

d=6+2*t

d=6+2t

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lyudmila [28]

Answer:

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25a-5b-8<br> ——————<br> 5a4b

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5a^2b^2+8. That would be ur answer

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3 years ago

Two types of coins are produced at a factory: a fair coin and a biased one that comes up heads 60 percent of the time. We have o

liraira [26]

Answer:

i) 0.1% probability that if the coin is actually fair, we reach a false conclusion.

ii) 0.05% probability that if the coin is actually unfair, we reach a false conclusion

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

Fair coin:

Comes up heads 50% of the time, so $p = 0.5$

1000 trials, so $n = 1000$

So

$E(X) = np = 1000*0.5 = 500$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.5*0.5} = 15.81$

If the coin lands on heads 550 or more times, then we shall conclude that it is a biased coin.

(i) If the coin is actually fair, what is the probability that we shall reach a false conclusion?

This is the probability that the number of heads is 550 or more, so this is 1 subtracted by the pvalue of Z when X = 549.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{549 - 500}{15.81}$

$Z = 3.1$

$Z = 3.1$ has a pvalue of 0.9990

1 - 0.9990 = 0.001

0.1% probability that if the coin is actually fair, we reach a false conclusion.

(ii) If the coin is actually unfair, what is the probability that we shall reach a false conclusion?

Comes up heads 60% of the time, so $p = 0.6$

1000 trials, so $n = 1000$

So

$E(X) = np = 1000*0.6 = 600$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.6*0.4} = 15.49$

If the coin lands on less than 550 times(that is, 549 or less), then we shall conclude that it is a biased coin.

So this is the pvalue of Z when X = 549.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{549 - 600}{15.49}$

$Z = -3.29$

$Z = -3.29$ has a pvalue of 0.0005

0.05% probability that if the coin is actually unfair, we reach a false conclusion

5 0

3 years ago

Need help asap I’m stuck

tiny-mole [99]

The product of the algebraic expressions $3a^{2}b^7$ and $5a^3b^8$ is $15a^5b^{15}$ . Hence, <u>3rd option</u> is the right choice.

<h3>How do we multiply two algebraic expressions?</h3>

When two algebraic expressions are multiplied, the like terms are to be combined and multiplied, and then they should be written as a single algebraic expression.

<h3>How do we solve the given question?</h3>

In the question, we are asked to find the product of the algebraic expressions $3a^{2}b^7$ and $5a^3b^8$ , that is:

( $3a^{2}b^7$ )( $5a^3b^8$ ).

To find the product of the two expressions, we combine the like terms:

$(3*5)(a^2*a^3)(b^7*b^8)$

= $(15)(a^5)(b^{15})$ (Using the law of exponents: $x^a*x^b = x^{a+b}$ )

Now we write them as a single expression: $15a^5b^{15}$ .

∴ The product of the algebraic expressions $3a^{2}b^7$ and $5a^3b^8$ is $15a^5b^{15}$ . Hence, <u>3rd option</u> is the right choice.

Learn more about the product of algebraic expressions at

brainly.com/question/4344214

#SPJ2

6 0

2 years ago

Read 2 more answers

Last year winter coat cost 175 this year the price increased 15% what is the new price

pogonyaev

Wow, finally a math one I know how to do :)

So, we take 175 and we add 15/100 or 15% to it.

The equation that we would use is 175(1.15)=cost

Your answer would be 201.25 after multiplying.

Hope this helped :)

(also, if you could, I'm like two brainliest's away from my next rank so can you...you know...) ;)

4 0

3 years ago