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3 years ago

PLZZZZ HELP MEEEEE IM GIVING 20 POINTS

1 answer:

3 0

1.theoretical is not counting the results of the experimentssince there are an equal number of red and black and red, the probblity (theoretical) of picking a black one is 1/3
experimental=number of outcomes happened/total number of tests4 times black, 4 tests, so 4/4 or 100%=experimental proablity

2. experimental considered previous trials and theoretical did not

3. theoretical=(1/4) the trials don't influence each other so1/(4*4)=1/16

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What expression shows 4 less than a number?<br> A. n + 4<br> B. n - 4<br> C. 4 - n<br> D. n(4)

Elis [28]

Answer:

Step-by-step explanation:

Because the expression in b is saying whatever number n is say 9 then your subtracting 4 by N or like I said take 9 for an example which=5

hope it helps

3 0

2 years ago

Read 2 more answers

Ray finds a log that is 90 inches long. How long is the log in feet and inches. Explain how you solved (1 yd= 3ft = 36 inches)

Shtirlitz [24]

Answer:

7.5

Step-by-step explanation:

1 foot is 12 inches so I just divided.

6 0

3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

3 years ago

What effect does replacing x with x + 2 have on the graph for the function f(x) ?

IrinaK [193]

The answer is A.

When replacing the variable 'x' by 'x + 2' the graph is shifted 2 units to the left. When we replace the variable, the function changes

from
$f(x) = |x - 3| + 2$

To
$f(x) = |(x + 2) - 3| + 2$
$f(x) = |x - 1| + 2$

And therefore the function is shifted 2 to the left.

3 0

3 years ago

solo savings bank received an initial deposit of $6000 it kept a percentage of this money in reserve based on the reserve rate a

iogann1982 [59]

Answer:

Reserve rate = 3%

Step-by-step explanation:

Reserve Ratio = Reserve Maintained with Central Bank / Deposit Liabilities × 100

Reserve maintained = $6,000

Deposit liabilities = $200,000

Reserve Rate = Reserve Maintained with Central Bank / Deposit Liabilities × 1000

=$6,000 / $200,000 × 100

=0.03 × 100

r=0.03 × 100

=3%

Therefore,

Reserve Rate =3%

8 0

4 years ago