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Kryger [21]
3 years ago
5

Beau made a new hamster house for Dr.cheeks out of 125.5 craft sticks. each stick weighs 0.5 ounce. How much does the hamster ho

use weigh?
Mathematics
2 answers:
andrezito [222]3 years ago
6 0

Answer:

62.75 ounces

Step-by-step explanation:


mafiozo [28]3 years ago
6 0

Answer:

62.75

Step-by-step explanation:

125.5×0.5=62.75

Brainliest please

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kupik [55]
First let's write out the inequality before choosing a graph.

x apples each weighing 1/3 of a pound: 1/3x

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So...

1/3x + y < 5

The maximum weight is 4 pounds since the total weight of both the grapes and apples are less than 5.

In the y-axis, the first, third, and fourth graphs already exceed the capacity of 5 pounds.

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Use the remainder theorem to find the remainder for (x ^5 + 32) divided by (x+2) and state whether or not the binomial is a fact
marta [7]

x+2 is a factor of x^5+32 because (by the remainder theorem) the remainder upon dividing x^5+32 by x+2 is (-2)^5+32=0.

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4 0
3 years ago
The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deduction
Len [333]

Answer:

(a) <em>                             </em><em>n</em> :      20           50          100         500

P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886    0.4444    0.5954    0.9376

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable <em>X</em> represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, <em>μ</em> = $16,642 and standard deviation is, <em>σ</em> = $2,400.

Assuming that the random variable <em>X </em>follows a normal distribution.

(a)

Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:

  • For a sample size of <em>n</em> = 20

P(\mu-200

                                           =P(-0.37

  • For a sample size of <em>n</em> = 50

P(\mu-200

                                           =P(-0.59

  • For a sample size of <em>n</em> = 100

P(\mu-200

                                           =P(-0.83

  • For a sample size of <em>n</em> = 500

P(\mu-200

                                           =P(-1.86

<em>                                  n</em> :      20           50          100         500

P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886    0.4444    0.5954    0.9376

(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample (\bar x) approaches the whole population mean (\mu_{x}).

Consider the probabilities computed in part (a).

As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.

So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

Thus, the correct option is (b).

8 0
3 years ago
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