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lozanna [386]
3 years ago
9

a country will produce 3000 missels this year. each of the short range missles costs $200,000 to produce. Each of the medium ran

ge missles cost $300,000 to produce. the ratio of the number of short range to medium range missles to long range missles the country plans to produce this year is 3:3:4. the country plans to use $870,000,000 to produce these missles. how much does it cost to produce longe range missles?
Mathematics
1 answer:
guapka [62]3 years ago
5 0

Answer:

Step-by-step explanation:

3000/10=300   300=10% of total missiles

300*3=900(30%)   That country (assuming it is Iran) would produce 900 short missiles and 900 medium missiles. 3000-900(2)=3000-1800=1200.

Iran will produce 1200 long range missiles.

For the cost of the missiles,

200,000*900=180,000,000. 300,000*900=270,000,000.

870M-(180M+270M)=870M-450M=420M

420,000,000/1200=350,000.

Long Range Missiles are $350,000 each... got me thinkin bout buying some...

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ABCD is a parallelogram. Segment AC = 4x + 10. Find the value of x and y.
valkas [14]

Answer:

The values of x and y in the diagonals of the parallelogram are x=0 and y=5

Step-by-step explanation:

Given that ABCD is a parallelogram

And segment AC=4x+10

From the figure we have the diagonals AC=3x+y and BD=2x+y

By the property of parallelogram the diagonals are congruent

∴ we can equate the diagonals AC=BD

That is 3x+y=2x+y

3x+y-(2x+y)=2x+y-(2x+y)

3x+y-2x-y=2x+y-2x-y

x+0=0 ( by adding the like terms )

∴ x=0

Given that segment AC=4x+10

Substitute x=0  we have AC=4(0)+10

=0+10

=10

∴ AC=10

Now (3x+y)+(2x+y)=10

5x+2y=10

Substitute x=0, 5(0)+2y=10

2y=10

y=\frac{10}{2}

∴ y=5

∴ the values of x and y are x=0 and y=5

5 0
4 years ago
Need your help please
erastovalidia [21]
I hope you can read cursive. Hope this helps!

7 0
4 years ago
Among the thirty largest U.S. cities, the mean one-way commute time to work is 25.8 minutes. The longest one-way travel time is
Dafna11 [192]

Answer:

A. 0.9015

B. 0.1658

C. 0.0132

Step-by-step explanation:

Given

Mean, μ of commuting time in New York is 39.7 minutes

Standard Deviation, σ = 7.5 minutes

Let x represent the commute time

For Normal Distribution, z = (x - μ) /σ

A. x is less than 30 minutes

P(x<30) = P(x - μ < 30 - μ)

P(x<30) = P((x - μ)/ σ < (30 - μ)/ σ)

P(x<30) = P(z < (30 - μ)/ σ)

P(x<30) = P(z < (30 - 39.7) / 7.5)

P(x<30) = P(z < (-9.7/7.5)

P(x<30) = P(z < -1.29)

P(x<30) = P(z > 1.29)

P(x<30) = 0.9015 -------- From z table

B. x is between 30 and 35

P(30>x<35) = P(30 -μ > x - μ < 35 - μ)

P(30>x<35) = P((30 -μ)/σ < (x - μ)/σ < (35 - μ) / σ)

P(30>x<35) = P((30 -μ)/σ < z < (35 - μ) / σ)

P(30>x<35) = P((30 -39.7)/7.5 < z < (35 - 39.7) / 7.5)

P(30>x<35) = P(-9.7/7.5 < z < -4.7/7.5)

P(30>x<35) = P(-1.29 < z < -0.63)

There are two points on the same side here; we calculate both probabilities and subtract to give

P(30>x<35) = P(-1.29 < z < 0) - P(0 < z < -0.63)

P(30>x<35) = P(0 < z < 1.29) - P(0 < z < 0.63)

P(30 > x < 35) = 0.9015 - 0.7357

P(30 > x < 35) = 0.1658

C. x is between 30 and 50

P(30>x<50) = P(30 -μ > x - μ < 50 - μ)

P(30>x<50) = P((30 -μ)/σ < (x - μ)/σ < (50 - μ) / σ)

P(30>x<50) = P((30 -μ)/σ < z < (50 - μ) / σ)

P(30>x<50) = P((30 -39.7)/7.5 < z < (50 - 39.7) / 7.5)

P(30>x<50) = P(-9.7/7.5 < z < 10.3/7.5)

P(30>x<50) = P(-1.29 < z < 1.37)

There are two points on different sides here; calculate both probabilities and add to give

P(30>x<50) = P(-1.29 < z < 0) + P(0 < z < 1.37)

P(30>x<50) = -P(0 < z < 1.29) + P(0 < z < 1.37)

P(30 > x < 50) = -0.9015 + 0.9147

P(30 > x < 50) = 0.0132

8 0
3 years ago
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Mamont248 [21]

Answer:

A

Step-by-step explanation:

Hope this helps

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5 0
4 years ago
Read 2 more answers
The lifetime of a certain type of battery is normally distributed with mean value 15 hours and standard deviation 1 hour. There
KIM [24]

Answer:

If the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 15

Standard Deviation, σ = 1

Sample size = 4

Total lifetime of 4 batteries = 40 hours

We are given that the distribution of lifetime is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling:

\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt4} = 0.5

We have to find the value of x such that the probability is 0.05

P(X > x)  = 0.05

P( X > x) = P( z > \displaystyle\frac{x - 40}{0.5})=0.05  

= 1 -P( z \leq \displaystyle\frac{x - 40}{0.5})=0.05  

=P( z \leq \displaystyle\frac{x - 40}{0.5})=0.95  

Calculation the value from standard normal z table, we have,  

\displaystyle\frac{x - 40}{0.5} = 1.64\\x = 40.825 \approx 40.83  

Hence, if the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.

8 0
4 years ago
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