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3 years ago

I have 3 times as many nickels as quarters. How many quarters do I have if I have

Mathematics

1 answer:

Nostrana [21]3 years ago

7 0

Answer:

18 Nickels, 6 Quarters

Step-by-step explanation:

For this problem, let's build a system of equations to find the number of nickels and dimes you have. Let N represent the number of nickels and Q represent the number of quarters you have.

The first equation, you have 3 times as many nickels as quarters:

N = 3Q

The second equation, you have a total of $2.40. Given we know nickels are $0.05 and quarters are $0.25:

0.05N + 0.25Q = 2.40

Now, let's use the two equations to find the number nickels and quarters you have.

N = 3Q

0.05N + 0.25Q = 2.40

0.05(3Q) + 0.25Q = 2.40

0.15Q + 0.25Q = 2.40

0.4Q = 2.40

Q = 6

Now, let's plug the value back into the equation:

N = 3Q

N = 3(6)

N = 18

And we can verify these values with the second equation:

0.05N + 0.25Q = 2.40

0.05(18) + 0.25(6) ?= 2.40

0.90 + 1.50 ?= 2.40

2.40 == 2.40

Therefore, we have found and verified the amount of Nickels and Quarters you have.

Cheers.

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3 years ago

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Answer:

B. 0.03110

Step-by-step explanation:

Given

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Required

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$P = 60\%$

Convert to decimal

$P = 0,6$

Next; Determine the Probability of Miss (q)

Opposite probabilities add up to 1;

So,

$p + q = 1$

$q = 1 - p$

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$q = 1 - 0.6$

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Next,is to determine the required probability;

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3 years ago

Consider the following hypothesis test.H0:μ1−μ2=0 Ha:μ1−μ2≠0The following results are for two independent samples taken from the

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Answer:

a) $z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53$

b) $p_v =2*P(z$

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Step-by-step explanation:

Information given

$\bar X_{1}= 104$ represent the mean for 1

$\bar X_{2}= 106$ represent the mean for 2

$\sigma_{1}= 8.4$ represent the population standard deviation for 1

$\sigma_{2}= 7.6$ represent the population standard deviation for 2

$n_{1}=80$ sample size for the group 1

$n_{2}=70$ sample size for the group 2

z would represent the statistic

Hypothesis to test

We want to check if the two means for this case are equal or not, the system of hypothesis would be:

H0: $\mu_{1}=\mu_{2}$

H1: $\mu_{1} \neq \mu_{2}$

The statistic would be given by:

$z =\frac{\bar X_1-\bar X_2}{\sqrt{\frac{\sigma^2_1^2}{n_1} +\frac{\sigma^2_2^2}{n_2}}}=$ (1)

Part a

Replacing we got:

$z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53$

Part b

The p value would be given by this probability:

$p_v =2*P(z$

Part c

Since the p value is higher than the significance level provided we have enogh evidence to FAIL to reject the null hypothesis and we can't conclude that the true means are different at 5% of significance

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Step-by-step explanation:

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