At what points does the helix r(t) = sin t, cos t, t intersect the sphere x2 + y2 + z2 = 65? (round your answers to three decima

At Central High School, 28% of the students are Freshmen, 26% are Sophomores, 24% are Juniors, and 22% are Seniors. The student

Jet001 [13]

Answer:

(D) This is a stratified random sample because a separate random sample is selected from each class

Step-by-step explanation:

A sample of size n is defined to be a stratified random sample if it is selected from a population which has been divided into a number of non overlapping groups or sub populations called strata, such that part of the sample is drawn at random from each stratum.

It is to be emphasized that good stratification requires that each of these strata should be internally homogeneous but externally should differ from one another.

The advantage of stratified sampling is low cost , greater accuracy and better coverage.

If we analyze the given scenario a sample is selected from each strata depicting its corresponding percentage in the population. It is internally homogeneous but externally different.

4 0

2 years ago

The hypotenuse of a right triangle is 50 and the short leg is 30, find the projection of the leg onto the hypotenuse

gladu [14]

Consider the attached figure. All the triangles shown are similar, so

CB/CA = CD/CB
30/50 = projection/30
projection = 30^2/50
projection = 18

6 0

2 years ago

Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production

Whitepunk [10]

Answer:

32% probability of a defect.

The expected number of defects for a 1000-unit production run is 320.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 10

Standard deviation = 0.15.

Probability of a defect

Weights less than 9.85 oz or greater than 10.15 oz will be classified as defects.

9.85 = 10 - 0.15

10.15 = 10 + 0.15

By the Empirical Rule, 68% of weights are between 9.85 and 10.15, that is, within the limits. 32% are not within the limits.

So there is a 32% probability of a defect.

Expected number of defects for a 1000-unit production run.

E(X) = 0.32*1000 = 320

The expected number of defects for a 1000-unit production run is 320.

5 0

2 years ago

For the binomial expansion of (x + y)^10, the value of k in the term 210x 6y k is a) 6 b) 4 c) 5 d) 7

Sloan [31]

Answer:

a) 6

Step-by-step explanation:

Expanding the polynomial using the formula:

$(x+y)^n=\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$

Also

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

I think you mean $210x^6y^4$

We can deduce that this term will be located somewhere in the middle. So I will calculate $k= 5; k=6 \text{ and } k =7$ .

For $k=5$

$\binom{10}{5} (y)^{10-5} (x)^{5}=\frac{10!}{(10-5)! 5!}(y)^{5} (x)^{5}= \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5! }{5! \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } \\ =\frac{30240}{120} =252 x^{5} y^{5}$

Note that we actually don't need to do all this process. There's no necessity to calculate the binomial, just $x^{n-k} y^k$

For $k=6$

$\binom{10}{6} \left(y\right)^{10-6} \left(x\right)^{6}=\frac{10!}{(10-6)! 6!}\left(y\right)^{4} \left(x\right)^{6}=210 x^{6} y^{4}$

5 0

2 years ago

Consider a regenerative vapor power cycle with two feedwater heaters, a closed one and an open one. Steam enters the first turbi

miss Akunina [59]

Answer:

Consider a regenerative vapor power cycle with two feedwater heaters, a closed one and an open one, and reheat. Steam enters the first turbine stage at 12 MPa, 480∘C, and expands to 2 MPa. Some steam is extracted at 2 MPa and fed to the closed feedwater heater. The remainder is reheated at 2 MPa to 440∘C and then expands through the second-stage turbine to 0.3 MPa, where an additional amount is extracted and fed into the open feedwater heater operating at 0.3 MPa. The steam expanding through the third-stage turbine exits at the condenser pressure of 6 kPa. Feedwater leaves the closed heater at 210∘C, 12 MPa, and condensate exiting as saturated liquid at 2 MPa is trapped into the open feedwater heater. Saturated liquid at 0.3 MPa leaves the open feedwater heater. Assume all pumps and turbine stages operate isentropically. Determine for the cycle

a. Draw the cycle on a T-S diagram using the same numbering in the schematic

b. Determine the thermal efficiency of the cycle.

c. Determine the mass flow rate of steam entering the first turbine of the cycle.

(i) Thermal efficiency of the cycle = 43.185 %

(ii) The mass flow rate of steam =93.66 kg/h

Step-by-step explanation:

So we have at

For Point 1 on the T-S diagram we have

p₁ = 80 bar,

t₁ = 480 °C,

From the super-heated steam tables we have

h₁ = 3349.6 kJ/kg, s₁ = 6.6613 kJ/kg·K

Point 2

p₂ = 20 bar

s₁ = s₂ =with x₂ = (6.6613 -6.6409)/(6.6849-6.6409) = 0.464

therefore h₂ =2953.1 + 0.464×(2977.1 - 2953.1) = 2964.22 kJ/kg

Point 3 on the T-S diagram we have

p₃ = 3 bar again s₁ = s₃ so we go to 3 bar on the steam tables and look up s = 6.6613 kJ/kg·K which is on the saturated steam tables

and x₃ is given as (6.6613 -1.6717)/(6.9916-1.6717) = 0.9379 and

h₃ = 561.43 + x₃×2163.5 = 2590.6 kJ/kg

Point 4

p₄ = 0.08 bar, s₁ = s₄, x₄ = 0.7949 and h₄ = 2083.45 kJ/kg

Point 5

p₅ = 0.08 bar, $h_{f5}$ = 173.84 kJ/kg

Point 6

Here h₆ is given by $h_{f5}$ plus the work done to move the water to the open heater therefore h₆ =

= 173.84 kJ/kg + 0.00100848×(3 - 0.08) × 100

= 173.84 kJ/kg + 0.29447616 kJ/kg = 174.13 kJ/kg

Point 7

p₇ = 3 bar, and $h_{f7}$ = 561.43 kJ/kg

Point 8

Here again work is done to convey the fluid t constant pressure thus

h₈ = $h_{f7} + v_{f7}$ × (p₈ - p₇)

561.43 kJ/kg + 0.00107317×(80 - 3)×100 = 569.69 kJ/kg

Point 9

p₉ = 80 bar and T₉ = 205°C

By interpolating the values on the subcooled teperature tables we get

x₉ = 0.5 and h₉ = 854.94 + 0.5× (899.79 - 854.94) = 877.365 kJ/kg

Point 10

p₁₀ = 20 bar, h₁₀ = $h_{f10}$ = 908.50 kJ/kg

point 11

Here h₁₁ = h₁₀ = 908.50 KJ/kg

For the closed feed water heater, energy and mass flow rate balance gives

m₁ × (h₂ - h₁₀) + (h₈ - h₉) = 0

Therefore m₁ = $\frac{ (h_{9} - h_8)}{(h_{2} - h_{10})}$ = 0.14967

while the open water heater we get

m₂×h₃+(1-m₁-m₂)×h₆+m₁×h₁₁ - h₇ = 0

from where m₂ = 0.11479

$W_{T}$ = (h₁-h₂) + (1 - m₁)(h₂ - h₃) +(1 - m₁ - m₂)(h₃ - h₄)

= 1076.11 kJ/kg

$W_{p}$ = (h₈ - h₇) + (1 - m₁ - m₂)×(h₆ - h₅)

= 8.4733 kJ/kg

Q = h₁ -h₉ = 2472.235 kJ/kg

Efficiency = η = $\frac{W_{T} - W_{P} }{Q}$ = 43.185 %

(ii) $W_{cycle} = m_1*(W_T -W_P)$

m'₁ = 100×10³/1066.63 = 93.66 kg/h

5 0

2 years ago