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2 years ago

Which is a better estimate for the volume of a cereal box? 1 cup 1 gallon

1 answer:

8 0

1 gallon. You’re estimating, so which one is about how much a box of cereal will hold? Well 1 cup is way to small, there are many cups in a box. A gallon is a better estimate.

You might be interested in

Which could be the first step in solving this system of equations by substitution?

Ainat [17]

Answer:

The first step is to replace the y in y-x=15 with 7x (we can do this because the first equation tells us that they're equal)

solve and get

(2.5,17.5)

or x= 2.5 y=17.5

Step-by-step explanation:

The first step is to recognize that y=7x which means that we can just replace the y in the second equation with 7x

7x-x=15

6x=15

x=2.5

Then we can solve for y

y=7(2.5)

y=17.5

3 0

3 years ago

5,25,125 find 10th term

anastassius [24]

Answer:

9765625

Step-by-step explanation:

There is a common ratio between consecutive terms , that is

r = 25 ÷ 5 = 125 ÷ 25 = 5

This indicates the sequence is geometric with n th term

$a_{n}$ = a₁ $r^{n-1}$

where a₁ is the first term and r the common ratio

Here a₁ = 5 and r = 5 , then

a₁₀ = 5 × $5^{9}$ = 5 × 1953125 = 9765625

6 0

3 years ago

Find the odds for and the odds against the event rolling a fair die and getting a 1 comma a 2 comma or a 5.

IrinaVladis [17]

Answer:

1/2 for or against

Step-by-step explanation:

There is a six sided die so the chance of getting one side is 1/6.

To get the sum for 3 sides, just do 1/6 + 1/6 + 1/6 = 3/6 or 1/2.

4 0

3 years ago

Solve for 5+n/8=4 show your work

Aleks [24]

Answer:

n = -8

Step-by-step explanation:

5+n/8=4

Subtract 5 from each side

5-5+n/8=4-5

n/8 = -1

Multiply each side by 8

n/8 * 8 = -1 *8

n = -8

5 0

3 years ago

The population of mosquitoes in a certain area increases at a rate proportional to the current pop-ulation, and in the absence o

ollegr [7]

Answer:

The population of mosquitoes in the area at any time t is:

$P(t)=201977.31-1977.31\times 2^{t}$

Step-by-step explanation:

The rate of growth of mosquitoes can be expressed as:

$\frac{dP}{dt}=kP$

$\frac{dP}{P}=k\ dt$

Integrate the above expression as follows:

$\int {\frac{dP}{P}} \, =\int {k\ dt} \, \\\ln|P|=kt+c\\e^{\ln|P|}=e^{kt+c}\\P=Ce^{kt}$

$\Rightarrow P=P_{0}e^{kt}$

It is provided that the population doubles every day.

Compute the value of k as follows:

$2=1\times e^{k\times1}\\2=e^{k}\\k=\ln (2)$

It is also provided that every day 20,000 mosquitoes are eaten.

The rate of growth per week can be expressed as:

$\frac{dP}{dt}=\ln(2)P-14000\\\frac{dP}{dt}-\ln(2)P=14000$

The integrating factor for this is:

$e^{\int {\ln(2)dP}}=e^{\ln(2)\int {dt}}=e^{\ln(2)t}$

Then,

$P(t)\ e^{-\ln(2)t}=\int {e^{-\ln(2)t}}-14000\, dt\\=-14000\int {e^{-\ln(2)t}}\, dt\\=-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C\\P(t)=(e^{-\ln(2)t})\times [-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C]\\=\frac{14000}{\ln(2)}+Ce^{-\ln(2)t}$

The initial population is 200,000.

Compute the value of C as follows:

$P(t)=\frac{140000}{\ln(2)}+Ce^{-\ln(2)t}\\200000=\frac{14000}{\ln(2)}+Ce^{-\ln(2)(0)}\\C=200000-\frac{140000}{\ln(2)}\\C=-1977.31$

Now substitute C in P (t),

$P(t)=\frac{140000}{\ln(2)}+Ce^{\ln(2)t}\\P(t)=201977.31-1977.31\times 2^{t}$

6 0

3 years ago