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Thepotemich [5.8K]
1 year ago
9

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

e are told Noah's ark was 300 cubits long, 50 cubits wide, and 30 cubits high. Estimate the volume of the ark (in cubic feet). Assume the ark has a shoe-box shape and that 1 palm = 3.10 inch.
Mathematics
1 answer:
olasank [31]1 year ago
6 0

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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Joe's Painting charges $100 plus $20 per hour to paint the exterior of your home. Steve's Painting charges $120 plus $15 per hou
Tanzania [10]

Joe's Painting: 20x + 100 = y

Steve's Painting: 15x + 120 = y

x = hours worked

y = total income

We can find when the two equations intersect by making them equal to each other. That means we put an equal sign in the middle. So, it would look something like this:

20x + 100 = 15x + 120

First, we have to move the 100 by subtracting it from both sides.

20x = 15x + 120 - (100)

20x = 15x + 20

Then, we need to move the 15x by subtracting it from both sides.

20 - (15x) = 20

5x = 20

Lastly, we need to divide 5 from both sides.

5x = 20/5

x = 4

Therefore, Joe and Steve would have to work for 4 hours in order for their models to be equal to each other.

5 0
2 years ago
Two random samples are taken from private and public universities
kati45 [8]

Answer:

Step-by-step explanation:

For private Institutions,

n = 20

Mean, x1 = (43120 + 28190 + 34490 + 20893 + 42984 + 34750 + 44897 + 32198 + 18432 + 33981 + 29498 + 31980 + 22764 + 54190 + 37756 + 30129 + 33980 + 47909 + 32200 + 38120)/20 = 34623.05

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (43120 - 34623.05)^2+ (28190 - 34623.05)^2 + (34490 - 34623.05)^2 + (20893 - 34623.05)^2 + (42984 - 34623.05)^2 + (34750 - 34623.05)^2 + (44897 - 34623.05)^2 + (32198 - 34623.05)^2 + (18432 - 34623.05)^2 + (33981 - 34623.05)^2 + (29498 - 34623.05)^2 + (31980 - 34623.05)^2 + (22764 - 34623.05)^2 + (54190 - 34623.05)^2 + (37756 - 34623.05)^2 + (30129 - 34623.05)^2 + (33980 - 34623.05)^2 + (47909 - 34623.05)^2 + (32200 - 34623.05)^2 + (38120 - 34623.05)^2 = 1527829234.95

Standard deviation = √(1527829234.95/20

s1 = 8740.22

For public Institutions,

n = 20

Mean, x2 = (25469 + 19450 + 18347 + 28560 + 32592 + 21871 + 24120 + 27450 + 29100 + 21870 + 22650 + 29143 + 25379 + 23450 + 23871 + 28745 + 30120 + 21190 + 21540 + 26346)/20 = 25063.15

Summation(x - mean)² = (25469 - 25063.15)^2+ (19450 - 25063.15)^2 + (18347 - 25063.15)^2 + (28560 - 25063.15)^2 + (32592 - 25063.15)^2 + (21871 - 25063.15)^2 + (24120 - 25063.15)^2 + (27450 - 25063.15)^2 + (29100 - 25063.15)^2 + (21870 - 25063.15)^2 + (22650 - 25063.15)^2 + (29143 - 25063.15)^2 + (25379 - 25063.15)^2 + (23450 - 25063.15)^2 + (23871 - 25063.15)^2 + (28745 - 25063.15)^2 + (30120 - 25063.15)^2 + (21190 - 25063.15)^2 + (21540 - 25063.15)^2 + (26346 - 25063.15)^2 = 1527829234.95

Standard deviation = √(283738188.55/20

s2 = 3766.55

This is a test of 2 independent groups. Let μ1 be the mean out-of-state tuition for private institutions and μ2 be the mean out-of-state tuition for public institutions.

The random variable is μ1 - μ2 = difference in the mean out-of-state tuition for private institutions and the mean out-of-state tuition for public institutions.

We would set up the hypothesis. The correct option is

-B. H0: μ1 = μ2 ; H1: μ1 > μ2

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

t = (34623.05 - 25063.15)/√(8740.22²/20 + 3766.55²/20)

t = 9559.9/2128.12528473889

t = 4.49

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [8740.22²/20 + 3766.55²/20]²/[(1/20 - 1)(8740.22²/20)² + (1/20 - 1)(3766.55²/20)²] = 20511091253953.727/794331719568.7114

df = 26

We would determine the probability value from the t test calculator. It becomes

p value = 0.000065

Since alpha, 0.01 > than the p value, 0.000065, then we would reject the null hypothesis. Therefore, at 1% significance level, the mean out-of-state tuition for private institutions is statistically significantly higher than public institutions.

4 0
2 years ago
A nutritionist attempts to determine an association between where food is prepared and the number of calories the food contains.
Oksi-84 [34.3K]
Given the conditional relative frequency table below which was generated by column using frequency table data comparing the number of calories in a meal to whether the meal was prepared at home or at a restaurant.

Number of Calories and Location of Meal Preparation.
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≥ 500 calories              0.15                      0.55                  0.28

< 500 calories              0.85                      0.45                  0.72

Total                             1.0                        1.0                    1.0

To determine whether there is an association between where food is prepared and the number of carories the food contain, we recall that an "association" exists between two categorical variables if the column conditional relative frequencies are different for the columns of the table. The bigger the differences in the conditional relative frequencies, the stronger the association between the variables. If the conditional relative frequencies are nearly equal for all categories, there may be no association between the variables. Such variables are said to be <span>independent.

For the given conditional relative freduency, we can see that there is a significant difference between the columns of the table.
i.e. 0.15 is significantly different from 0.55 and 0.85 is significantly different from 0.45

Therefore, we can conclude from the given answer options that t</span><span>here is an association because the value 0.15 is not similar to the value 0.55</span>
8 0
2 years ago
Box A is 567 inches high, 678 inches wide, and 789 inches long. It has a maximum capacity of 1,200 marbles, Box B has three time
kotykmax [81]

Answer:

1701 high

2712 wide

789 long

= 4413

Step-by-step explanation:

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2 years ago
The orbit of the planet Venus is nearly circular. An astronomer develops a model for the orbit in which the sun has coordinates
klio [65]
Since the Venus orbits round the sun, the sun is the center of the circular path of the revolution of the planet, Venus.

Thus, the distance of the planet, Venus fron the sun is given by the distance between the points (0, 0) and (41, 53).

Recall that the distance between two points (x_1, y_1) and (x_2, y_2) is given by d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Thus, the distance between the points (0, 0) and (41, 53) is given by:

d= \sqrt{(41-0)^2+(53-0)^2}  \\  \\ = \sqrt{41^2+53^2} = \sqrt{1,681+2,809}  \\  \\ = \sqrt{4,490} =67 \ units

Given that each unit of the plane represents 1 million miles, therefore, the distance from the sun to the Venus is 67 million miles.
8 0
1 year ago
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