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White raven [17]

3 years ago

13

Parsons Bank offers two checking-account plans. The No Frills plan charges 25 cents per check whereas the Simple Checking plan

costs $6 per month plus 5 cents per check. For what number of checks per month will the Simple Checking plan cost less?

1 answer:

deff fn [24]3 years ago

4 0

No frills = $0.25(x)
simple = $6 + $0.05(x)
set equal to each other
0.025(x) = 6 + 0.05(x)
rearrange like terms together
0.025(x) - 0.05(x) = 6
0.020(x) = 6
x =300
Simple plan will be cheaper once you use more than 300 checks

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A bread machine is marked "reduced by 15%, now only $101.15".

Art [367]

(100-15=85%)
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4 years ago

Maci and I are making a small kite. Two sides are 10". Two sides are 5". The shorter diagonal is 6". Round all your answers to t

Art [367]

Answer:

A. 4".

B. Approximately 9.54".

C. Approximately 13.54".

Step-by-step explanation:

Please find the attachment.

Let x be the distance from the peak of the kite to the intersection of the diagonals and y be the distance from the peak of the kite to the intersection of the diagonals.

We have been given that two sides of a kite are 10 inches and two sides are 5 inches. The shorter diagonal is 6 inches.

A. Since we know that the diagonals of a kite are perpendicular and one diagonal (the main diagonal) is the perpendicular bisector of the shorter diagonal.

We can see from our attachment that point O is the intersection of both diagonals. In triangle AOD the side length AD will be hypotenuse and side length DO will be one leg.

We can find the value of x using Pythagorean theorem as:

$(AO)^2=(AD)^2-(DO)^2$

$x^{2}=5^2-3^2$

$x^{2}=25-9$

$x^{2}=16$

Upon taking square root of both sides of our equation we will get,

$x=\sqrt{16}$

$x=\pm 4$

Since distance can not be negative, therefore, the distance from the peak of the kite to the intersection of the diagonals is 4 inches.

B. We can see from our attachment that point O is the intersection of both diagonals. In triangle DOC the side length DC will be hypotenuse and side length DO will be one leg.

We can find the value of y using Pythagorean theorem as:

$(OC)^2=(DC)^2-(DO)^2$

Upon substituting our given values we will get,

$y^2=10^2-3^2$

$y^2=100-9$

$y^2=91$

Upon taking square root of both sides of our equation we will get,

$y=\sqrt{91}$

$y\pm 9.539392$

$y\pm\approx 9.54$

Since distance can not be negative, therefore, the distance from intersection of the diagonals to the top of the tail is approximately 9.54 inches.

C. We can see from our diagram that the length of longer diagram will be the sum of x and y.

$\text{The length of the longer diagonal}=x+y$

$\text{The length of the longer diagonal}=4+9.54$

$\text{The length of the longer diagonal}=13.54$

Therefore, the length of longer diagonal is approximately 13.54 inches.

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3 years ago

Use the balanced scale to find the conversion factor that can be used to convert the number of blocks to the weight of the block

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Find Total weight at left side

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Now

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3 years ago

United Flight 15 from New York's JFK airport to San Francisco uses a Boeing 757-200 with 182 seats. Because some people with res

Tcecarenko [31]

Answer:

There is a 29.27% probability that the flight is overbooked. This is not an unusually low probability. So it does seem too high so that changes must be made to make it lower.

Step-by-step explanation:

For each passenger, there are only two outcomes possible. Either they show up for the flight, or they do not show up. This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

A probability is said to be unusually low if it is lower than 5%.

For this problem, we have that:

There are 200 reservations, so $n = 200$ .

A passenger consists in a passenger not showing up. There is a .0995 probability that a passenger with a reservation will not show up for the flight. So $\pi = 0.0995$ .

Find the probability that when 200 reservations are accepted for United Flight 15, there are more passengers showing up than there are seats available.

X is the number of passengers that do not show up. It needs to be at least 18 for the flight not being overbooked. So we want to find $P(X < 18)$ , with $\pi = 0.0995, n = 200$ . We can use a binomial probability calculator, and we find that:

$P(X < 18) = 0.2927$ .

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AlexFokin [52]

Answer:

1 ten.

Step-by-step explanation:

1 ten = 10 ones/units

7 0

3 years ago

Read 2 more answers