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3 years ago

10

Victor has 1 and a half hours before he has to leave. He starts watching a movie that runs for 107 minutes. How many minutes of

the movie will he miss?

1 answer:

Phoenix [80]3 years ago

6 0

He does not miss any of the movie

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PLEASE ANSWER ASAP WILL MARK BRAINLIEST

katrin2010 [14]

Answer:

48 hats and 104 shirts

Step-by-step explanation:

These are the equations you build from the problem:

h + s = 152

8.50h + 12s = 1656

This is how I solved them:

s= 152-h

8.5h + 12(152-h) = 1656

8.5h + 1824 - 12h = 1656

Solve for h

h= 48

Put this into first equation (h +s = 152) to get s

8 0

3 years ago

Read 2 more answers

Plz help me with this math and also explain

jasenka [17]

Step-by-step explanation:

<h2>[1]</h2>

SI = $250
Rate (R) = 12 $\sf \dfrac{1}{2}$ %
Time (t) = 4 years

$\longrightarrow \tt { SI = \dfrac{PRT}{100} } \\$

$\longrightarrow \tt { 250 = \dfrac{P \times 12\cfrac{1}{2} \times 4}{100} } \\$

$\longrightarrow \tt { 250 = \dfrac{P \times \cfrac{25}{2} \times 4}{100} } \\$

$\longrightarrow \tt { 250 = \dfrac{P \times 25 \times 2}{100} } \\$

$\longrightarrow \tt { 250 = \dfrac{P \times 50}{100} } \\$

$\longrightarrow \tt { 250 \times 100 = P \times 50} \\$

$\longrightarrow \tt { 25000 = P \times 50} \\$

$\longrightarrow \tt { \dfrac{25000}{50} = P } \\$

$\longrightarrow \underline{\boxed{ \green{ \tt { \$ \; 500 = P }}}} \\$

Therefore principal is $500.

<h2>__________________</h2>

<h2>[2]</h2>

2/7 of the balls are red.
3/5 of the balls are blue.
Rest are yellow.
Number of yellow balls = 36

Let the total number of balls be x.

→ Red balls + Blue balls + Yellow balls = Total number of balls

$\longrightarrow \tt{ \dfrac{2}{7}x + \dfrac{3}{5}x + 36 = x} \\$

$\longrightarrow \tt{ \dfrac{10x + 21x + 1260}{35} = x} \\$

$\longrightarrow \tt{ \dfrac{31x + 1260}{35} = x} \\$

$\longrightarrow \tt{ 31x + 1260= 35x} \\$

$\longrightarrow \tt{ 1260= 35x-31x} \\$

$\longrightarrow \tt{ 1260= 4x} \\$

$\longrightarrow \tt{ \dfrac{1260 }{4}= x} \\$

$\longrightarrow \underline{\boxed{ \tt { 315 = x }}} \\$

Total number of balls is 315.

A/Q,

3/5 of the balls are blue.

$\longrightarrow \tt{ Balls_{(Blue)} =\dfrac{3 }{5}x} \\$

$\longrightarrow \tt{ Balls_{(Blue)} =\dfrac{3 }{5}(315)} \\$

$\longrightarrow \tt{ Balls_{(Blue)} = 3(63)} \\$

$\longrightarrow \underline{\boxed{ \green {\tt { Balls_{(Blue)} = 189 }}}} \\$

8 0

3 years ago

If 4 (x-2/3) = -18, what is the value of 2x

BaLLatris [955]

Answer:

-11.5

Step-by-step explanation:

4(x-2/3)=-18

4x-8/3=-18

4x-8=-18×3

4x-8=-54

4x=-54+8

4x=-46

x=-46/4

x=-11.5

4 0

3 years ago

The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) for Determine

salantis [7]

The question is incomplete. Here is the complete question.

The probability density function of the time to failure of an electronic component in a copier (in hours) is

$f(x)=\frac{e^{\frac{-x}{1000} }}{1000}$

for x > 0. Determine the probability that

a. A component lasts more than 3000 hours before failure.

b. A componenet fails in the interval from 1000 to 2000 hours.

c. A component fails before 1000 hours.

d. Determine the number of hours at which 10% of all components have failed.

Answer: a. P(x>3000) = 0.5

b. P(1000<x<2000) = 0.2325

c. P(x<1000) = 0.6321

d. 105.4 hours

Step-by-step explanation: <em>Probability Density Function</em> is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.

So, probability function is given by:

P(a<x<b) = $\int\limits^b_a {P(x)} \, dx$

Then, for the electronic component, probability will be:

P(a<x<b) = $\int\limits^b_a {\frac{e^{\frac{-x}{1000} }}{1000} } \, dx$

P(a<x<b) = $\frac{1000}{1000}.e^{\frac{-x}{1000} }$

P(a<x<b) = $e^{\frac{-b}{1000} }-e^\frac{-a}{1000}$

a. For a component to last more than 3000 hours:

P(3000<x<∞) = $e^{\frac{-3000}{1000} }-e^\frac{-a}{1000}$

Exponential equation to the infinity tends to zero, so:

P(3000<x<∞) = $e^{-3}$

P(3000<x<∞) = 0.05

There is a probability of 5% of a component to last more than 3000 hours.

b. Probability between 1000 and 2000 hours:

P(1000<x<2000) = $e^{\frac{-2000}{1000} }-e^\frac{-1000}{1000}$

P(1000<x<2000) = $e^{-2}-e^{-1}$

P(1000<x<2000) = 0.2325

There is a probability of 23.25% of failure in that interval.

c. Probability of failing between 0 and 1000 hours:

P(0<x<1000) = $e^{\frac{-1000}{1000} }-e^\frac{-0}{1000}$

P(0<x<1000) = $e^{-1}-1$

P(0<x<1000) = 0.6321

There is a probability of 63.21% of failing before 1000 hours.

d. P(x) = $e^{\frac{-b}{1000} }-e^\frac{-a}{1000}$

0.1 = $1-e^\frac{-x}{1000}$

$-e^{\frac{-x}{1000} }=-0.9$

${\frac{-x}{1000} }=ln0.9$

-x = -1000.ln(0.9)

x = 105.4

10% of the components will have failed at 105.4 hours.

5 0

4 years ago

Is that better for you guys please hurry

prohojiy [21]

The answer to the question

7 0

3 years ago