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charle [14.2K]
2 years ago
6

Six less than a number in algebraic form

Mathematics
1 answer:
Novay_Z [31]2 years ago
4 0

Answer:

six less than a number is n-6

Step-by-step explanation:

n-6

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Sue has $2.10 in dimes and nickels. If she has 12 more dimes than nickels, how many of each coin does she have?
Andreas93 [3]
Sue would have 21 dimes and 12 less of 21 is 9, so sue has 9 nickels

dimes: 21
nickels: 9
8 0
3 years ago
 plz help need help fast The figure is the net for a rectangular prism.
Art [367]
Two small rectangle at the end areas are:
3x5=15
15x2=30
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Length:5+3+5+3=16
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30+144=174
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3 years ago
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7 0
2 years ago
Jimmy wants to order pizza for the birthday party on Thursday each pizza has four slices if the eight kids at the party each eat
blagie [28]

Answer: 14 Pizzas

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8 0
3 years ago
A random sample of 747 obituaries published recently in Salt Lake City newspapers revealed that 344 (or 46%) of the decedents di
cupoosta [38]

Answer:

The probability value is almost equal to 0. Implying that the proportion of people dying  in that particular interval if deaths occurred randomly throughout the year is unusual.

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of decedents who died in the three-month period following their birthdays.

A random sample of 747 obituaries published recently in Salt Lake City newspapers revealed that 344 (or 46%) of the decedents died in the three-month period following their birthdays (123).

The probability (p) of anyone dying in any quarter if people die randomly during the year is simply 0.25.

The random variable <em>X</em> follows a Binomial distribution with parameters n = 747 and p = 0.25.

But the sample selected is too large and the probability of success is small.

So a Normal approximation to binomial can be applied to approximate the distribution of <em>p</em> if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=747\times 0.25=186.75>10\\n(1-p)=747\times (1-0.46)=560.25>10

Thus, a Normal approximation to binomial can be applied.

So,  p\sim N(\hat p,\ \frac{\hat p(1-\hat p)}{n}).

Compute the probability that 46% or more would die in that particular interval if deaths occurred randomly throughout the year as follows:

P (p\geq 0.46)=P(\frac{p-\hat p}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}>\frac{0.46-0.25}{\sqrt{\frac{0.25(1-0.25)}{747}}})

                   =P(Z>13.25)\\=1-P(Z

 *Use a <em>z</em> table for the probability.

The probability value is almost equal to 0. This probability is very low indicating that the proportion of people dying  in that particular interval if deaths occurred randomly throughout the year is unusual.

3 0
3 years ago
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