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2 years ago

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The solid glass award that Jillian was given has a total height of 15 inches as shown. A rectangular pyramid with a base of 6 in

ches by 4 inches and a height of 15 inches. A rectangular prism with a length of 6 inches, width of 4 inches, and height of 2 inches. What is the total volume of glass used to make the award? 120 inches cubed 152 inches cubed 168 inches cubed 204 inches cubed

2 answers:

ludmilkaskok [199]2 years ago

5 0

Answer:

204inches i think

Step-by-step explanation:

makkiz [27]2 years ago

3 0

Answer:

it's a 120

Step-by-step explanation:

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For each 3 mm of coloured fabric Alex uses to make his curtains, he also uses 2 cm of white fabric. Express the amount of white

Leokris [45]

Answer: 2cm/.2cm

Step-by-step explanation:

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2 years ago

The rate of change in revenue for Under Armour from 2004 through 2009 can be modeled by dR /dt = 13.897t + 284.653 t where R is

MakcuM [25]

Answer:

The revenue is $R(9) = \$ 876.9$

Step-by-step explanation:

From the question we are told that

The rate of change in revenue for Under Armour from 2004 through 2009 is

$\frac{d R }{dt } = 13.897t + \frac{284.653}{t}$

Now

$dR = (13.897t + \frac{284.653}{t})dt$

Integrating both sides to obtain R(t)

$\int\limits dR = \int\limits (13.897t + \frac{284.653}{t})dt$

$\int\limits dR = \int\limits (13.897\frac{t^2}{2} + 284.653(ln (t)) ) + C$

=> $R(t) = 6.9485t^2 + 284.653(ln (t) ) + C$

From the question we have that at t = 8 $R(8)= \$ 725.2 \ million$

$725.2 = 6.9485(8)^2 + 284.653(ln (8) ) + C$

=> $C = -311.4$

So

$R(t) = 6.9485t^2 + 284.653(ln (t) ) -311.4$

At t = 9

$R(9) = 6.9485*(9)^2 + 284.653(ln (9) ) -311.4$

$R(9) = \$ 876.9$

7 0

2 years ago

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a

Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q = -t/100 + c

$Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c} = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}$

when t = 0, Q = 200 L × 1 g/L = 200 g

$Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}$

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

$2 = 200e^{(-t/100)}\\\frac{2}{200} = e^{(-t/100)}$

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

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2 years ago

Which number produces a rational number when multiplied by 0.5

ivanzaharov [21]

10 is your answer because it is a terminating or repeating decimals

5 0

2 years ago

Read 2 more answers

the time taken by a student to the university has been shown to be normally distributed with mean of 16 minutes and standard dev

Naya [18.7K]

Answer:

a) 2.84% probability that he is late for his first lecture.

b) 5.112 days

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 16, \sigma = 2.1$

a. Find the probability that he is late for his first lecture.

This is the probability that he takes more than 20 minutes to walk, which is 1 subtracted by the pvalue of Z when X = 20. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20 - 16}{2.1}$

$Z = 1.905$

$Z = 1.905$ has a pvalue of 0.9716

1 - 0.9716 = 0.0284

2.84% probability that he is late for his first lecture.

b. Find the number of days per year he is likely to be late for his first lecture.

Each day, 2.84% probability that he is late for his first lecture.

Out of 180

0.0284*180 = 5.112 days

4 0

2 years ago