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suter [353]
3 years ago
10

Allison is making bracelets to sell in a school craft show. Each bracelet takes her 12 minutes to make, and she's been making br

acelets for 120 minutes. She can spend no more than 300 minutes making bracelets. If x represents the remaining number of bracelets that Allison can make, which inequality represents the situation?

Mathematics
2 answers:
PSYCHO15rus [73]3 years ago
6 0
Let
x-------------> <span>represents the remaining number of bracelets that Allison can make
</span>
we know that
1) <span>for the first 120 minutes Allison can make
120/12=10 bracelets

</span>2) <span>for the next 180 minutes Allison can make 
180/12=15 bracelets</span>

therefore
remaining minutes=300-120-------> 180 minutes
so
12x \leq 180  -----------> this <span>inequality represents the situation
</span>
Ghella [55]3 years ago
4 0

Answer:

The answer is, 12x + 120 -< 300

Step-by-step explanation:

For the Plato users: This situation can be represented by a linear inequality. The time it takes to make one bracelet, 12 minutes, is the rate of change, which is multiplied by the remaining number of bracelets that Allison can make, x. The time she has already spent making bracelets, 120 minutes, is constant.  

Allison cannot spend more than 300 minutes making bracelets, so the less than or equal to sign is used in the inequality. The correct inequality is shown below.

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<span>The missing value that represents the rate for Marie is 1/60. As mentioned in the table, rate (r) is the part (p) per minute (t). Let's check if it works for Lea. If p = 1/80t, then r = (1/80t)/t = 1/80 * t/t = 1/80 * 1 = 1/80. So, the formula works. Now, calculate the rate for Marie. If p = 1/60t, then r = (1/60t)/t = 1/60 * t/t = 1/60 * 1 = 1/60</span>
7 0
3 years ago
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Suppose cattle in a large herd have a mean weight of 1217lbs1217 lbs and a variance of 10,40410,404. What is the probability tha
AlexFokin [52]

Answer:

0.2460 = 24.60% probability that the mean weight of the sample of cows would differ from the population mean by more than 11 lbs if 116 cows are sampled at random from the herd.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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

In a set with mean \mu and standard deviation(which is the square root of the variance) \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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 1217, \sigma = \sqrt{10414} = 102, n = 116, s = \frac{102}{\sqrt{116}} = 9.475

What is the probability that the mean weight of the sample of cows would differ from the population mean by more than 11 lbs if 116 cows are sampled at random from the herd?

This is 2 multiplied by the pvalue of Z when X = 1217 - 11 = 1206. So

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{1206 - 1217}{9.475}

Z = -1.16

Z = -1.16 has a pvalue of 0.1230

2*0.1230 = 0.2460

0.2460 = 24.60% probability that the mean weight of the sample of cows would differ from the population mean by more than 11 lbs if 116 cows are sampled at random from the herd.

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3 years ago
Explain one way to find the area of the polygon at the right. Then find the area in square units.
kari74 [83]

Answer:

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3 years ago
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Five cards are dealt from a standard 52-card deck. (a) What is the probability that we draw 1 ace, 1 two, 1 three, 1 four, and 1
Rudik [331]

Answer:

Step-by-step explanation:

As there are total 52 cards in a deck and we have to draw a set of 5 cards, we can use the formula of combination to find the total number of possible ways of drawing 5 cards.

Number of ways to draw 5 cards = N_T

N_T\;=\;({}^NC_k)\\\\N_T\;=\;({}^{52}C_5)\\\\N_T\;=\;2,598,960

(a) Assuming the cards are drawn in order (would not affect the probability). The of getting Ace, 2, 3, 4 and 5 can be obtained by multiplying the probability of getting cards below 6 (20/52) with the probability of getting 5 different cards (4 choices for each card).

P(a)\;=\;\frac{20}{52}*\frac{4}{52}*\frac{4}{51}*\frac{4}{50}*\frac{4}{49}*\frac{4}{48}\\\\P(a)\;=\; 1.3133*10^{-6}

(b) For a straight we require our set to be in a sequence. The choices for lowest value card to produce a sequence are ace, 2, 3, 4, 5, 6, 7, 8, 9, or 10. Hence, the number of ways are ({}^{10}C_1).

For each card we can draw from any of the 4 sets. It can be described mathematically as: ({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)\;=\;[({}^{4}C_1)^5]

Therefore, the total outcomes for drawing straight are:

N_S\;=\;({}^{4}C_1)*({}^{4}C_1)^5\;=\;10240

Thus, the probability of getting a straight hand is:

P(b)\;=\;\frac{N_S}{N_T}\\\\P(b)\;=\;\frac{10240}{2598960}\\\\P(b)\;=\; 0.0039

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3 years ago
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Law Incorporation [45]

Answer:

Step-by-step explanation:

13*3=39ft (if thats what ur looking for)

6 0
3 years ago
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