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2 years ago

Blocking in experimental design is useful because it:

Mathematics

2 answers:

umka21 [38]2 years ago

8 0

Answer:

Experimental design is the process of attaining specified objectives through a properly planned study. Planning an experiment properly is very important because it will give the right type of data and a sufficient sample size. Blocking or randomization is useful because it controls for multiple variables in an experiment.

MrMuchimi2 years ago

7 0

The correct awnser is B. APEX

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Sandra just finished planting avacados,

Elena L [17]

Answer:392.5

Step-by-step explanation:

50/2=25

25*25=625

625*3.14=1962.5

1962.5/5=392.5

8 0

2 years ago

Katie needs to measure water for a recipe she is using to cook dinner. she should measure the liquid in

Lisa [10]

The answer to that question is B. pints

7 0

3 years ago

Read 2 more answers

Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8

Delvig [45]

We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51$

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98$

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34$
Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86$

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.

3 0

3 years ago

Una caja de 25 newtons se suspende mediante un cable con diámetro de 2cm¿cuál es el esfuerzo aplicado al cable?

Naddik [55]

El cable experimenta un esfuerzo axial de 79577.472 pascales por el peso de la caja.

<h3>¿Cómo calcular el esfuerzo aplicado sobre el cable?</h3>

La caja tiene masa y está sometida a un campo gravitacional, por tanto, tiene un peso (W), en newtons. Por el principio de acción y reacción (tercera ley de Newton), encontramos que el cable es tensionado debido a ese peso y su área transversal experimenta un esfuerzo axial (σ), en pascales.

Asumiendo una distribución uniforme de la fuerza sobre toda la superficie transversal de la cuerda, tenemos que el esfuerzo axial se calcula mediante la siguiente expresión:

σ = W / (π · D² / 4)

Donde:

W - Peso de la caja, en newtons.
D - Diámetro del área transversal de la caja, en metros.

Si sabemos que W = 25 N y D = 0.02 m, entonces el esfuerzo axial aplicado a la cuerda es:

σ = 25 N / [π · (0.02 m)² / 4]

σ ≈ 79577.472 Pa

<h3>Observación</h3>

La falta de problemas verificados en español sobre esfuerzos axiales obliga a buscar uno equivalente en inglés.

Para aprender más sobre esfuerzos axiales: brainly.com/question/13683145

#SPJ1

5 0

1 year ago

Can someone help me solve this?

atroni [7]

Answer:

Step-by-step explanation:

speed=d÷t.

s=1/4=1/20

cross multiply your equation

6 0

2 years ago