All categories

3 years ago

What is the volume of 4ft by 6ft by 1ft by 2ft by 8ft?

Mathematics

2 answers:

Nonamiya [84]3 years ago

4 0

Calculate this:

V = (4 ft)(6 ft)(2.67 ft)

= (24 ft^2)(2.67 ft) = 64.08 ft^3

bija089 [108]3 years ago

4 0

V = (4 ft)(6 ft)(2.67 ft)
V = (24 ft^2)(2.67 ft) = 64.08 ft^3

You might be interested in

Emma is standing 10 feet away from the base of a tree and tries to measure the angle of elevation to the top. She is unable to g

Vaselesa [24]

The complete question in the attached figure

step 1
find the tree height for an angle of elevation of 55°
tan 55°=opposite side angle 55°/adjacent side angle 55°
opposite side angle 55°=the tree height h1
adjacent side angle 55°=10 ft
so
tan 55°=h/10-------> h=10*tan 55°-----> h=14.28 ft

step 2
find the tree height for an angle of elevation of 75°
tan 75°=opposite side angle 75°/adjacent side angle 75°
opposite side angle 75°=the tree height h2
adjacent side angle 75°=10 ft
so
tan 75°=h2/10-------> h2=10*tan 75°-----> h2=37.32 ft

therefore

the height of the tree is within the range [14.26,37.32]

Part 1) 4.2 ft

the value is not in the interval [14.26,37.32]
therefore
it is Not Reasonable

Part 2) 14.7 ft

the value is in the interval [14.26,37.32]
therefore
it is Reasonable

Part 3) 24.4 ft

the value is in the interval [14.26,37.32]
therefore
it is Reasonable

Part 4) 33.9 ft

the value is in the interval [14.26,37.32]
therefore
it is Reasonable

Part 5) 39.1ft

the value is not in the interval [14.26,37.32]
therefore
it is Not Reasonable

Part 6) 58.7 ft

the value is not in the interval [14.26,37.32]
therefore
it is Not Reasonable

4 0

4 years ago

A rectangular garden has length twice as great as its width. A second rectangular garden has the same length as the first garden

Slav-nsk [51]

I say 12 because that would mean the first rectangle is 6x12 and the second is 10x12

4 0

4 years ago

What is the probability of not picking a red face card when you draw a card at random from a pack of 52 cards?

aniked [119]

Answer:

3/13

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The

TiliK225 [7]

Correct question:

A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The company ships them to restaurants in boxes of 9 salmon, to grocery stores in cartons of 16 salmon, and to discount outlet stores in pallets of 64 salmon. To forecast costs, the shipping department needs to estimate the standard deviation of the mean weight of the salmon in each type of shipment. Complete parts (a) and (b) below.

a. Find the standard deviations of the mean weight of the salmon in each type of shipment.

b. The distribution of the salmon weights turns out to be skewed to the high end. Would the distribution of shipping weights be better characterized by a Normal model for the boxes or pallets?

Answer:

Given:

Mean, u = 44

Sd = 3

The company ships in boxes of 9, cartons of 16 and pallets of 64.

a) For the standard deviations of the mean weight of the salmon in each type of shipment, lets use the formula: $\frac{s.d}{\sqrt{u}}$

i) For the standard deviation of the mean weight of salmon in boxes of 9, we have:

$\frac{s.d}{\sqrt{u}}$

$= \frac{3}{\sqrt{9}}$

$= \frac{3}{3} = 1$

The standard deviation = 1

ii) For the standard deviation of the mean weight of salmon in cartons of 16, we have:

$\frac{s.d}{\sqrt{u}}$

$= \frac{3}{\sqrt{16}}$

$= \frac{3}{4} = 0.75$

Standard deviation = 0.75

iii) For the standard deviation of the mean weight of salmon in pellets of 64, we have:

$\frac{s.d}{\sqrt{u}}$

$= \frac{3}{\sqrt{64}}$

$= \frac{3}{8} = 0.375$

Standard deviation = 0.375

b) The distribution of shipping weights would be better characterized by a Normal model for the pallets, because regardless of the underlying distribution, the sampling distribution of the mean approaches the Normal model as the sample increases.

5 0

4 years ago

lamars bowling scores were 124, 150, and 161, during league competition. if lamars previous bowling average was a 129, how much

Darina [25.2K]

43 is the correct answer

8 0

3 years ago

Read 2 more answers