Please help I’ll mark you as brainliest if correct!

Step-by-step explanation:

120=2/5t

120×5/2=t

t=600/2

t=300

hope it helps u

Answer:

(1,-7)

Step-by-step explanation:

1) (5,-4). 4 units to the left will be (1,-4)

2) (1,-4). 3 units down will be (1,-7)

Answer:

A. I

Step-by-step explanation:

The other person is wrong and I took the test

Answer:

±6

Step-by-step explanation:

Given the mathematical expression;

√36 < √42 < √49

But, √36 = 6 * 6 = 6²

√49 = 7 * 7 = 7²

Simplifying further, we would substitute the new values respectively;

±[6] < √42 < ±[7]

Therefore, evaluating √36 is equal to ±6.

Answer:

Answer:

The wright answer is A

Step-by-step explanation:

[tex] \sqrt{20} = 4.47 \\ \frac{14}{3} = 4.67 \\ 4.47 < 4.5 < 4.67[/tex]

Answer:

See Annex ( Tree diagram )

Step-by-step explanation:

1) Attached

2) Probability of the second ball is yellow is

P(2nd ball is yellow ) = Probability 2nd ball s yellow (if the first one was black) + Probability 2nd ball s yellow (if the first one was Yellow)

P(2nd ball is yellow ) = 0,204 + 0,045

P(2nd ball is yellow ) = 0,249

Answer: B. 183 cm^3

V = pi r^2 h/3

V = pi (5^2) (7/3)

V = pi (25) (2.33)

V = 183

Answer:

B.183 [tex]cm^{3}[/tex]

Step-by-step explanation:

Volume of cone is [tex]V=\pi r^{2} \frac{h}{3}[/tex]

where h=7 and r=5 (because r=d/2, where d=10)

V=[tex]5^{2}\frac{7}{3}\pi[/tex]

V=25·[tex]\frac{7}{3} \pi[/tex]

V=175/3[tex]\pi[/tex]

V=183.2595715 or 183 [tex]cm^{3}[/tex]

Answer:

95°

Step-by-step explanation:

95°

They are vertical angles.

Answer:

95 degree write it fast jakakkaoaoa

Answer: Midpoint: (2, 1)

Step-by-step explanation:

Midpoint formula: ((x1 + x2)/2 , (y1 + y2)/2)

(-4 + 8)/2 = 2

(-8 + 10)/2 = 1

Answer: (2, 1)

Answer:

x5

Step-by-step explanation:

Answer: C.

Step-by-step explanation:

EDGE 2023

Answer:

here you go............

Step-by-step explanation:

I hope it helped

Answer:

D.

Step-by-step explanation:

just done the assignment

Linear regression line is a algebraic model to show the relationship between the two models by putting the value of the one variable to get the value of other.. For every 1 pound of nitrogen added to the field, the amount of corn yielded increases by 0.43 bushels. Therefore the option B is the correct option.

The equation of the experiment to know the relationship between the amount of nitrogen added to a cornfield and the number of bushels of corn produced is,

[tex]y=0.43x+28.5[/tex]

Linear regression line is a algebraic model to show the relationship between the two models by putting the value of the one variable to get the value of other.

Linear regression line can be represent as,

[tex]y=Ax+B[/tex]

Here x is the explanatory variable and y is the dependent variable. A is the slope of the line and B is the intercept.

In the given equation there are two variable given.

Variable x represent the amount of nitrogen added, and the variable y, represents the number of bushels of corn produced.

As putting the value of x the y increases with value of number 0.43. Thus as we are increasing the value of the nitrogen with unit 1 the number of bushels of corn produced is increased by unit 0.43.

Hence for every 1 pound of nitrogen added to the field, the amount of corn yielded increases by 0.43 bushels. Therefore the option B is the correct option.

Learn more about the linear regression here;

https://brainly.com/question/25311696

Answer: 27

Step-by-step explanation:

Answer:

d

Step-by-step explanation:

Answer:

its d

Step-by-step explanation:

Answer:

a) The total engine volume is 30.3265 in³

b) the mass of air contained in the engine is 0.00129 lbm

c) The efficiency of the engine is 55.6%

Step-by-step explanation:

Given the data in the question;

The configuration is in line with four-stroke, four-cylinder engine.

Bore ( d ) = 3.17 in

Stroke ( L ) = 3.48 in

compression ratio ( r ) = 10.6 : 1

First, we calculate the swept Volume V₁ = π/4 × d²L

we substitute

V₁ = π/4 × (3.17 )² × (3.48) = 27.4655 in³

Now, Compression Ration; r = Vs + Vc / Vc

r = (V₁ + V₂) / V₂

we substitute

10.6 = (27.4655 + V₂) / V₂

10.6V₂ = 27.4655 + V₂

10.6V₂ - V₂ = 27.4655

9.6V₂ = 27.4655

V₂ = 2.861 in³

So

a) total engine volume

[tex]V_{BDC[/tex] = V₁ + V₂

we substitute

[tex]V_{BDC[/tex] = 27.4655 in³ + 2.861 in³

[tex]V_{BDC[/tex] = 27.4655 in³ + 2.861 in³

[tex]V_{BDC[/tex] = 30.3265 in³

Therefore, The total engine volume is 30.3265 in³

b) The mass of air

given that; T = 540°R, P = 14.7 psia

we know that; PV = mRT   { R is constant }

we solve for m

m = PV / RT

we substitute

m = (14.7 × 30.3265)  / ( 0.3704 × 1728 )540

m = 445.79955 / 345627.648

m = 0.00129 lbm

Therefore, the mass of air contained in the engine is 0.00129 lbm

c) The efficiency

from table { properties of air}

At 540°R; u₁ = 92.04 BTu/lbm, u[tex]_r[/tex]₁ = 144.32

Now process 1 - 2 ( Isentropic )   u[tex]_r[/tex]₁ / u[tex]_r[/tex]₂  = V₁/V₁ = r

so,  u[tex]_r[/tex]₁ / u[tex]_r[/tex]₂ = r

144.32  / u[tex]_r[/tex]₂ = 10.6

u[tex]_r[/tex]₂ = 144.32/10.6

u[tex]_r[/tex]₂ = 13.61

Thus, at u[tex]_r[/tex]₂ = 13.61, u₂ = 235 BTU/lbm, T₂ = 1340°R

Now, we know that;

P₁V₁/T₁ = P₂V₂/T₂  ⇒ (P₁/V₁)r = P₂/T₂

⇒ ( 14.7 / 540 )10.6 = P₂/1340

P₂ = 386.664 psia  

process 2 - 3; constant volume heat addition;

Now, At T₃ = 3600°R, u₃ = 721.44 BTU/lbm, u[tex]_r[/tex]₃ = 0.6449

so, [tex]q_{in[/tex] =  u₃ -  u₂

we substitute

so, [tex]q_{in[/tex] =  721.44  -  235  = 486.44 BTU/lbm

Next, process 3 - 4; Isentropic  

u[tex]_r[/tex]₄/u[tex]_r[/tex]₃ = V₄/V₃ = r

u[tex]_r[/tex]₄ / 0.6449  = 10.6

u[tex]_r[/tex]₄ = 6.8359

At u[tex]_r[/tex]₄ = 6.8359; u₄ = 308 BTU/lbm

∴

Heat Rejection [tex]q_{out[/tex] = u₄ - u₁ = 308 - 92.04 = 215.96 BTU/lbm

so, [tex]W_{net[/tex] = [tex]q_{in[/tex] - [tex]q_{out[/tex]

[tex]W_{net[/tex] = 486.44 - 215.96

[tex]W_{net[/tex] = 270.48 BTU/lbm

so the efficiency; η[tex]_{thermal[/tex] = ([tex]W_{net[/tex] / [tex]q_{in[/tex] ) × 100%

η[tex]_{thermal[/tex] = ( 270.48 / 486.44  ) × 100%

η[tex]_{thermal[/tex] = ( 0.55604  ) × 100%    

η[tex]_{thermal[/tex] = 55.6%

Therefore, The efficiency of the engine is 55.6%

Answer:

Option B and D are correct

Step-by-step explanation:

Exponential functions are really useful in the real world situation. They can be used to solve following

a) Population Models

b) Determination of area and perimeters

c) Determine time related things such as half life, time of happening of an event etc.

d) Useful for solving financial problems such as computing investments etc.

Hence, option B and D

Answer:

Step-by-step explanation:

Use SOH CAH TOA to recall how the trig functions fit on a triangle

SOH: Sin(Ф)= Opp / Hyp

CAH: Cos(Ф)= Adj / Hyp

TOA: Tan(Ф) = Opp / Adj

Let's use CAH  1st  , on the Lenovo building,  b/c we  have they Hypotenuse (Hyp)  and an angle and we want to know the other side of the Lenovo building

Cos(30) = x/294

294*Cos(30) = x

254.611 = x

Sin(30) = y / 294

294*Sin(30) = y

147 = y

perimeter = 147+254.611+294

perimeter Lenovo = 695.611 feet

For Asus building

hyp = sqrt [ ( 144.5*[tex]\sqrt{2}[/tex])^2 +  144.5*[tex]\sqrt{2}[/tex])^2 )]

hyp = 289

perimeter = 289+144.5*[tex]\sqrt{2}[/tex]+144.5*[tex]\sqrt{2}[/tex]

perimeter Asus = 697.7077

The difference of the two buildings is

697.7077 -  695.611 =2.09671 feet

the difference is 2.10 feet ( rounded to nearest 100th)

Answer:

Part a:

Since the calculated z= 8.1649 falls in the critical region  z < ± 1.28 we reject H0 and conclude that the assembly time using the new method is faster.

Part b:

P (Type II Error)=  0.6628

Step-by-step explanation:

The null and alternate hypothesis are

H0: μ1 ≥ μ2   against the claim Ha: μ1 < μ2

This is left tailed test

The critical region for this test at 0.10 ∝ is z < ± 1.28

The test statistic

z= x1-x2/ s/√n

z= 66-61/ 3/√24

z= 5/0.61238

z= 8.16486

Since the calculated z= 8.1649 falls in the critical region  z < ± 1.28 we reject H0 and conclude that the assembly time using the new method is faster.

Part b:

P (Type II Error)  = P ( accept H0/ H0 is false)

z ( ∝)= X-u/s/√n

1.28= X-61/0.61238

X= 61.7838

So I will incorrectly fail to reject the null as long as a draw a sample mean that greater than 61.7838

 The probability of a Type II error is given by

P (z> 61.7838-66/24)

P (z>- 0.42161)

From the tables

P (z>- 0.42161)=  0.6628

Answer:

56 m^2

Explanation: L=2, H=2, W=2, d = 6.63324958 m, TSA = 56 m^2, LSA = 48 m^2, TSA = 4 m^2, BSA = 4 m^2, V = 24 m^3

Answer:

0.3372 = 33.72% probability of having a sample mean of 116.8 or less for a random sample of this size

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean IQ score of its members is 118, with a standard deviation of 17.

This means that [tex]\mu = 118, \sigma = 17[/tex]

Sample of 35:

This means that [tex]n = 35, s = \frac{17}{\sqrt{35}}[/tex]

What is the probability of having a sample mean of 116.8 or less for a random sample of this size?

This is the pvalue of Z when X = 116.8. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{116.8 - 118}{\frac{17}{\sqrt{35}}}[/tex]

[tex]Z = -0.42[/tex]

[tex]Z = -0.42[/tex] has a pvalue of 0.3372

0.3372 = 33.72% probability of having a sample mean of 116.8 or less for a random sample of this size

Answer:

umm to me its 2--%456

Step-by-step explanation:

Answer:

a-As the only factor of interest is type of electric motor, thus instead of using 1-factor randomization, a randomized block design is used where at maximum one motor of each type is tested every day.

b- The test statistic is given as

[tex]F_{I-1,(I-1)(J-1)}=\dfrac{12\left(\Sigma^{5}_{i=1}\bar{X}^{2}_{i.}-IJ\bar{X}^2_{..}\right) }{\Sigma^5_{i=1}\Sigma^4_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}[/tex]

Step-by-step explanation:

a-

This is done such that the days are considered as blocks and the randomization is only occuring within the block. The order of testing is random. However the condition is implemented such that one motor of each type is tested each day.

b-

The F-Value is given as

[tex]F-Value=\dfrac{MSA}{MSAB}[/tex]

Here MSA is given as

[tex]MSA=\dfrac{J\Sigma^I_{i=1}(\bar{X}_{i.}-\bar{X}_{..})^2}{I-1}[/tex]

Here

Similarly MSAB is given as

[tex]MSAB=\dfrac{\Sigma^I_{i=1}\Sigma^J_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}{(I-1)(J-1)}[/tex]

Here

By putting these values and simplifying, equation becomes:

[tex]F-Value=\dfrac{MSA}{MSAB}\\\\F-Value=\dfrac{\dfrac{J\Sigma^I_{i=1}(\bar{X}_{i.}-\bar{X}_{..})^2}{I-1}}{\dfrac{\Sigma^I_{i=1}\Sigma^J_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}{(I-1)(J-1)}}\\\\F-Value=\dfrac{\dfrac{4\Sigma^5_{i=1}(\bar{X}_{i.}-\bar{X}_{..})^2}{5-1}}{\dfrac{\Sigma^5_{i=1}\Sigma^4_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}{(5-1)(4-1)}}\\\\[/tex]

This is further simplified as

[tex]F-Value=\dfrac{\dfrac{4\Sigma^5_{i=1}(\bar{X}_{i.}-\bar{X}_{..})^2}{4}}{\dfrac{\Sigma^5_{i=1}\Sigma^4_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}{12}}\\\\F-Value=\dfrac{\Sigma^5_{i=1}(\bar{X}_{i.}-\bar{X}_{..})^2}{\dfrac{\Sigma^5_{i=1}\Sigma^4_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}{12}}\\\\[/tex]

The numerator can be written as

[tex]F-Value=\dfrac{\Sigma^{5}_{i=1}\bar{X}^{2}_{i.}-IJ\bar{X}^2_{..}}{\dfrac{\Sigma^5_{i=1}\Sigma^4_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}{12}}\\\\F-Value=\dfrac{12(\Sigma^{5}_{i=1}\bar{X}^{2}_{i.}-IJ\bar{X}^2_{..})}{{\Sigma^5_{i=1}\Sigma^4_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}}\\[/tex]

Answer:

the area is 72

Step-by-step explanation:

Answer:

See attached

Step-by-step explanation:

Review the picture

Answer:

There are 20 odd numbers between 10 and 50

Step-by-step explanation:

Here are all the odd numbers! Hope this helps!

11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49

[tex]\huge{\textbf{\textsf{{\color{navy}{An}}{\purple{sw}}{\pink{er}} {\color{pink}{:}}}}}[/tex]

C. 4.