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2 years ago

7

The LaSalle High School senior class raised funds for an end of the year cruise getaway. The city gave them a special package fo

r the port fees and taxes - a flat fee of $1,500 for the entire class's port fees and taxes. It costs $367 per student to board the cruise. If your class raised $16,600, how many students can attend?

1 answer:

Brilliant_brown [7]2 years ago

3 0

Knowns:
C = mx + b
m = 367
b = 1500

C = 367x + 1500

Cost (c) = 16,600

16600 = 367x + 1500
subtract 1500 from both sides

15,100 = 367x
Divide both sides by 367

x = 41.14

41 Students can attend the trip! Hope this helps

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Sara, Mark and Caroline share £102 in a ratio 1:3:2. How much money does each person get?

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Answer:

Sara- 17

Mark- 51

Caroline- 34

Step-by-step explanation:

1) start by adding the ratios together, so 1+3+2= 6. Due to the order of the ratio, Sara, gets 1 art of the money, Mark gets 3 parts and Caroline get 2 parts.

2) take the total amount of money (102) and divide it by (1+3+2). 102/6 is 17.

3) divide the money by the amount each person gets. Sara only has one part, so she has £17, Mark gets 3 parts so he gets £51, and Caroline gets 2 parts, so she ends up with £34.

4) do 17+34+51 to check it equals 102. ( it does ;D)

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2 years ago

A basket contains 14 white eggs, 15 brown eggs, and 11 lemons. Taylor is in a hurry to make breakfast and picks something from t

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Okay so probability is just percentage of a whole, right?

So you have 14 White Eggs + 15 Brown Eggs + 11 Lemons.

Add all those numbers together and you get your whole.

14 + 15 = 29 29+11 = 40

40 is your whole.

So because you want to know how likely it is to pick up an egg, you would follow these steps.

100/40 = 2.5 (For each part of the 40, it is worth 2.5 percent.)

2.5 x 29 = 72.5

Your probability of picking an egg out of the bask is 72.5 percent or 72.5 out of 100.

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2 years ago

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2 years ago

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A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = <u><em>the length of the calls, in minutes.</em></u>

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = <u>0.1271</u>

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

2 years ago

The cubit is an ancient unit. Its length equals six palms. (A palm varies from 2.5 to 3.5 inches depending on the individual.) W

olasank [31]

Assuminhg the ark has a shoe-box (cuboid) shape it´s volume would be:

$V_{cuboid}=lenght*width*height$

We have this three measurements ( $l=300cubits;w=50cubits;h=30cubits)$ ) and it can be as simple as replacing them in the equation and solving but they are in all in $cubits$ . We will convert them to $ft$ because the problem requires the answer in $ft^{3}$ . In order to do this we will use the given equivalences:

$1cubit=6palms\\1palm=3.10in$

and another one:

$1ft=12in$

First we will convert from cubits to palms:

$l=300cubits*\frac{6palms}{1cubit}=1800palms\\w=50cubits*\frac{6palms}{1cubit}=300palms\\h=30cubits*\frac{6palms}{1cubit}=180palms\\$

now from $palms$ to $in$ :

$l=1800palms*\frac{3.10in}{1palm}=5580in\\w=300palms*\frac{3.10in}{1palm}=930in\\h=180palms*\frac{3.10in}{1palm}=558in\\$

now from $in$ to $ft$ :

$l=5580in*\frac{1ft}{12in}=465ft\\w=930in*\frac{1ft}{12in}=77.5ft\\h=558in*\frac{1ft}{12in}=46.5ft$

We can calculate the Volume now like this:

$V_{ark}=465ft*77.5ft*46.5ft=1675743.75ft^{3}$

The volume of trhe ark would be $1675743.75ft^{3}$

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2 years ago