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3 years ago

credit cards place a 3 digit security code on the back of cards what is the probability that a code starts with a number 7

1 answer:

8 0

This answer is not 3/4. The answer is 1/10. I found this by knowing that there are 10 numbers and there is a 1/10 chance that it will start with 7. I would really appreciate it if I could get brainliest. Thanks!! =D

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A drawing shows 4 apples and 6 bananas. Which is the ratio of bananas to apples? 6:4 3:2 6 4 all of the above

ioda

<span> 4 apples and 6 bananas

so ration bananas to apples = 6:4 or 3:2 or 6 4

answer is

</span><span>all of the above</span>

8 0

3 years ago

Read 2 more answers

The 18th term of an arithmetic sequence is 49 and the first term is 15.

Tju [1.3M]

Answer:

Step-by-step explanation:

49-15=34

34/17=2

For this sequence, you can write this function:

f(x)=15+2(x-1)

Check:

f(18)=15+2(18-1)

f(18)=15+2(17)

f(18)=15+34

f(18)=49

So:

f(10)=15+2(10-1)

f(10)=15+2(9)

f(10)=15+18

f(10)=33

5 0

3 years ago

Which is bigger? four liters or one gallon?

hjlf

4 liters is slightly bigger

5 0

3 years ago

Which form most quickly reveals the vertex?

lana [24]

Answer:

first option

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

m(x) = - 2 (x - 6)² + 18 ← is in vertex form

with vertex = (6, 18 )

7 0

3 years ago

Read 2 more answers

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

1 year ago