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4 years ago

7

Miranda needs to cut strips of paper that are 1/8 of an inch. If she has piece of paper that is 7 and 3/4 inches long. How many

strips of paper can she make? *

1 answer:

Sloan [31]4 years ago

8 0

Answer:

62

if you turn 1/8 and 7 3/4 to decimals they turn into 1/8=0.125 and 7 3/4=7.75 and 7.75/0.125=62

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Alyssa, Bart and Meghan are

adoni [48]

Hi,

Essentially, we are saying that Alyssa gets 5 parts of the whole thing (140,000), Bart gets 3 parts of the whole and Meghan gets 2 parts of the whole. The sum total of their parts = the whole. So now we need to find out how much one part is. Thus,"one part" should be the variable. Substitute p for "one part", so Alyssa gets 5p, Bart gets 3p, and Meghan gets 2p. Since the sum of the parts = the whole, your problem sets up as: 5p + 3p + 2p = 140,000 --> 10p = 140,000 --> 10p/10 = 140,000/10 --> x = 14,000.

Alyssa gets 5p or (5 x 14,000)

Bart gets 3p or (3 x 14,000)

Meghan gets 2p or (2 x 14,000)

If the whole was a loss, it would be a negative, and instead of getting money, they would lose money.

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3 years ago

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Dmitry_Shevchenko [17]

(r^10)-2 would be the expression

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4 years ago

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Nimfa-mama [501]

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b or c

Step-by-step explanation:

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3 years ago

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2 years ago

A random sample of 100 people from City A has an average IQ of 120 with a SD of 18. Independently of this, a random sample of 15

Sloan [31]

Answer:

$z=\frac{(120-116)-0}{\sqrt{\frac{18^2}{100}+\frac{15^2}{150}}}}=1.837$

$p_v =P(z>1.837)=1-P(Z$

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the average IQ on city A is signficantly higher than city B at 5% of singificance.

Step-by-step explanation:

$\bar X_{A}=120$ represent the mean for sample 1

$\bar X_{B}=116$ represent the mean for sample 2

$s_{A}=18$ represent the sample standard deviation for 1

$s_{B}=15$ represent the sample standard deviation for 2

$n_{A}=100$ sample size for the group 2

$n_{B}=150$ sample size for the group 2

$\alpha$ Significance level provided

z would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if residents of City A smarter on average, the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{B}\leq 0$

Alternative hypothesis: $\mu_{A} - \mu_{B}> 0$

We don't have the population standard deviation's, but the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{A}-\bar X_{B})-\Delta}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$z=\frac{(120-116)-0}{\sqrt{\frac{18^2}{100}+\frac{15^2}{150}}}}=1.837$

P value

Since is a one right tailed test the p value would be:

$p_v =P(z>1.837)=1-P(Z$

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the average IQ on city A is signficantly higher than city B at 5% of singificance.

5 0

3 years ago