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3 years ago

6

a video store charges a one-time membership fee of $12.00 plus $1.50 per video rental. How many videos can stewart rent if he sp

ends $21?

1 answer:

Natasha_Volkova [10]3 years ago

4 0

He could rent 6 videos if he spends $21.00

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The there angles in a triangle ge in the ratio

Serggg [28]

Answer:

Step-by-step explanation:

6 0

2 years ago

2/3* (1/3+3 1/4 help pls now

ale4655 [162]

Answer:

43/18

Step-by-step explanation:

2/3(1/3+ 3 1/4)

First solve parenthesis.

2/3(1/3+ 3 1/4)

find the lowest common denominator (12) to add the two fractions.

1/3= 4/12;

3 1/4= 13/4 = 39/12

4/12+ 39/12= 43/12

2/3(43/12)

Then multiply the fractions.

2/3(43/12)

Numerator: 2*43= 86

Denominator: 3*12= 36

86/36

Last simplify your fraction.

43/18 or 2 7/18

4 0

3 years ago

According to Internet security experts, approximately 90% of all e-mail messages are spam (unsolicited commercial e-mail), while

KonstantinChe [14]

Answer:

0.0656

Step-by-step explanation:

For each message, we have these following probabilities:

90% probability it is spam.

10% probability it is legitimate.

Compute the probability that the first legitimate e-mail she finds is the fifth message she checks:

The first four all spam, each with a 90% probability.

The fifth legitimate, with a 10% probability.

$P = (0.9)^{4} \times 0.1 = 0.0656$

3 0

3 years ago

A cone has a height of 7ft and a radius of 4ft which equation can find the volume of the cone v=1.3 (7.2)

Margaret [11]

Answer:

The volume of a cone is:

V=(hπr^2)/3, where h=height and r=radius

V=(7π4^2)/3

V=112π/3 ft^3

V≈117.29 ft^3 (to nearest hundredth of a cubic foot)

Step-by-step explanation:

4 0

3 years ago

The half-life of the isotope Osmium-183 is 12 hours. Choose the equation below that gives the remaining mass of Osmium-183 in gr

raketka [301]

The given equations are incomprehensible, I'm afraid...

You're given that osmium-183 has a half-life of 12 hours, so for some initial mass <em>M</em>₀, the mass after 12 hours is half that:

1/2 <em>M</em>₀ = <em>M</em>₀ exp(12<em>k</em>)

for some decay constant <em>k</em>. Solve for this <em>k</em> :

1/2 = exp(12<em>k</em>)

ln(1/2) = 12<em>k</em>

<em>k</em> = 1/12 ln(1/2) = - ln(2)/12

Now for some starting mass <em>M</em>₀, the mass <em>M</em> remaining after time <em>t</em> is given by

<em>M</em> = <em>M</em>₀ exp(<em>kt </em>)

So if <em>M</em>₀ = 590 g and <em>t</em> = 36 h, plugging these into the equation with the previously determined value of <em>k</em> gives

<em>M</em> = 590 exp(36<em>k</em>) = 73.75

so 73.75 ≈ 74 g of Os-183 are left.

Alternatively, notice that the given time period of 36 hours is simply 3 times the half-life of 12 hours, so 1/2³ = 1/8 of the starting amount of Os-183 is left:

590/8 = 73.75 ≈ 74

6 0

3 years ago