A manufacturing company has two assembly lines, one with old equipment andone with new. The old assembly line generates products at a rate of 48 per hour,
the new one at 60 per hour. If both assembly lines are operating at the same
time, how long will it take them to create a total of 180 products?

Answers

Answer 1
Answer:

Using unitary method, the manufacturing company will take 1 hour 40 minutes to create a total of 180 products.

What is unitary method?

The unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value.

Number of products generated by old assembly line in 60 minutes = 48

Number of products generated by old assembly line in 1 minute = 48/60 = 0.8

Number of products generated by new assembly line in 60 minutes = 60

Number of products generated by new assembly line in 1 minutes = 60/60 = 1

Total production in 1 minute = 1 + 0.8 = 1.8

Time taken to produce 180 products = 180/1.8 = 100 minutes = 1 hour 40 minutes

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Answer 2
Answer:

Answer:

the first one would take 3 hours and 45 mins to make 180 by itself

the second one would take 3 hours to make 180 by its self

if this ant the answer let me  know


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Please help fast!!! I need help with question attached!!

Answers

Answer:

26 degrees

Step-by-step explanation:

2 angles must equal to 180 of they share the same side/line so,

180-154=26

15. Simplify sin(90° - O) cos 0 - sin(180° +0) sin e.​

Answers

Answer:  cos²(θ) + sin(θ)sin(e)

Step-by-step explanation:

sin (90° - θ)cos(Ф) - sin(180° + θ) sin(e)

Note the following identities:

sin (90° - θ) = cos(x)

sin (180° + θ) = -sin(x)

Substitute those identities into the expression:

   cos(x)cos(x)  -  -sin(x)sin(e)

=  cos²(x) + sin(x)sin(e)

HELP HELP 20 Points

Geometry

Answers

270° → (3π)/(2) ≈ 4.7

360° → 2π ≈ 6.3

60° → (π)/(3) ≈ 1.1

45° → (π)/(4) ≈ 0.8

135° → (3π)/(4) ≈ 2.4

225° → (5π)/(4) ≈ 3.9

What are the solutions of this quadratic equation?
x2 − 10x = -34

Answers

The roots of the equation are imaginary numbers, 5+6i, 5-6i.

What is a Quadratic Equation?

A quadratic equation is what can be written in the form of ax²+bx+c=0

The quadratic equation is

x² -10x = -34

x² -10x +34 = 0

The roots formula for a quadratic equation is given by

x = ( -b \pm √(b^2 -4ac))/(2a)

Here a = 1

b = -10

c =  34

The roots of the equation are

x = ( 10 \pm √((-10)^2 -4*34))/(2)

\rm x = ( 10 \pm √(-36))/(2)\n\n\nx = 5 \pm 6i

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Here it is, step by step:

x^2 - 10x = -34
x^2 - 10x + 34 = 0
We cannot easily factor anything, so we resort to the quadratic equation. The work is attached as an image.

Can someone help me pls

Answers

Answer:

27/36

Step-by-step explanation:

Count the white region (which is 36), and then count the shaded region (which is 27). And write the shaded  region on the upper part (numerator), and the white part on the lower portion (denominator).

One question in the survey asked how much time per year the children spent in volunteer activities. The sample mean was 14.76 hours and the sample standard deviation was 16.54 hours.Required:

a. Based on the reported sample mean and sample standard deviation, explain why it is not reasonable to think that the distribution of volunteer times for the population of South Korean middle school students is approximately normal.
b. The sample size was not given in the paper, but the sample size was described as large. Suppose that the sample size was 500. Explain why it is reasonable to use a one-sample t confidence interval to estimate the population mean even though the population distribution is not approximately normal.
c. Calculate and interpret a confidence interval for the mean number of hours spent in volunteer activities per year for South Korean middle school children.

Answers

Answer:

a. If the distribution was normal, many values would be negative, what is incompatible with the response variable (hours dedicated to volunteer activities).

b. If the sample is big, accordingly to the Central Limit Theorem, the sampling distribution shape tends to be normally-like, so we can apply a one-sample t-test.

c. The 95% confidence interval for the mean is (13.307, 16.213).

Step-by-step explanation:

a. If the distribution was normal, the values with one or more standard deviation below the mean would be negative, what is incoherent for this case. This, in a normal distribution, represents approximately 16% of the values.

If we calculate the probabilty for a normal distribution with the sample parameters, the probability of having "negative hours" is 18.6% (see picture attached).

b. If the sample is big, accordingly to the Central Limit Theorem, the sampling distribution shape tends to be normally-like, so we can apply a one-sample t-test.

The sampling distribution standard deviation is also reduced by a factor of 1/√n.

c. We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=14.76.

The sample size is N=500.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=(s)/(√(N))=(16.54)/(√(500))=(16.54)/(22.3607)=0.7397

The t-value for a 95% confidence interval is t=1.965.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=1.965 \cdot 0.7397=1.453

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 14.76-1.453=13.307\n\nUL=M+t \cdot s_M = 14.76+1.453=16.213

The 95% confidence interval for the mean is (13.307, 16.213).