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Lina20
10 days ago
15

A survey conducted five years ago by the health center at a university showed that 18% of the students smoked at the time. This

year a new survey was conducted on a random sample of 200 students from this university, and it was found that 50 of them smoke. We want to find if these data provide convincing evidence to suggest that the percentage of students who smoke has changed over the last five years. What are the test statistic (Z) and p-value of the test?
Mathematics
1 answer:
tester [8.8K]10 days ago
5 0
The test statistic (Z) is 2.5767, and the p-value of the test is 0.009975. The null hypothesis suggests that the smoking rate among students has not changed, while the alternative indicates otherwise. The z-statistic for the sampled proportion is computed, yielding z ≈ 2.5767. As we investigate whether the smoking percentage has shifted over the preceding five years, the two-tailed p-value is found to be 0.009975. This result is significant at a 99% confidence level, demonstrating substantial evidence that the percentage of smoking students has changed.
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Crane Company reports the following for the month of June.
babunello [8412]

Respuesta:

Crane Company

Informes financieros de junio

a) Costo de bienes disponibles para la venta = $5,250

b) Costo unitario de promedio móvil para:

i) 1 de junio: = $5

ii) 12:  = $4.75

iii) 15: = $4.75

iv) 23:  = $5.75

v) 27:  = $5.25

Explicación paso a paso:

a) Cálculos:

Fecha Explicación Unidades Costo Unitario Costo Total Costo Promedio Móvil

1 de junio Inventario 150 $4 $600 $4.000

12 Compra 450 5 2,250 4.750

15 Venta 500 7 3,500 4.750

23 Compra 400 6 2,400 5.750

27 Venta 420 8 3,360 5.250

30 Inventario 80

Costo de bienes disponibles para la venta = Costo del Inventario Inicial + Costo de Compras = $5,250 + ($600 + 2,250 + 2,400)

b) Costo unitario de promedio móvil para:

i) 1 de junio: Costo de bienes disponibles/Unidades disponibles = $5 ($600/150)

ii)  12: Costo de bienes disponibles/Unidades disponibles = $4.75 ($600 + 2,250/600)

iii)  15: Costo de bienes disponibles/Unidades disponibles = $4.75 ($475/100)

iv)  23: Costo de bienes disponibles/Unidades disponibles = $5.75 ($475 + 2,400)/500

v)  27: Costo de bienes disponibles/Unidades disponibles = $5.25 ($420/80)

8 0
2 days ago
Which number can each term of the equation be multiplied by to eliminate the fractions before solving? m – negative StartFractio
zzz [9080]

Answer: To remove fractions prior to solving, each term in the equation must be multiplied by 4.

Step-by-step explanation:

Consider the given expression:

-\frac{3}{4}m-\frac{1}{2}=2+\frac{1}{4}m

It is essential to simplify this before attempting to solve it.

Since the denominators differ, identifying the Least Common Denominator (LCD) is necessary.

Break down the denominators into their prime components:

4=2*2=2^2\\2=2

Select 2^2, as it possesses the greatest exponent. Thus:

LCD=2^2=4

Ultimately, to remove the fractions before solving, multiply both sides by 4:

(4)(-\frac{3}{4}m)-(4)(\frac{1}{2})=(4)(2)+(4)(\frac{1}{4}m)\\\\-3m-2=8+m

7 0
23 days ago
Three classes of school children are selling tickets to the school play. the number of tickets sold by these classes, and the nu
lawyer [9240]
Aqe4hjwrs4jyjreajrsjwtj

6 0
1 month ago
Read 2 more answers
When jumping, a flea rapidly extends its legs, reaching a takeoff speed of 1.0 m/s over a distance of 0.50 mm . part a what is t
zzz [9080]

The flea experiences an acceleration of 1,000 m/s².

Detailed explanation

This scenario involves motion with constant acceleration.

The relevant variables include the following.

\boxed{u \ or \ v_i = initial \ velocity}

\boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}

\boxed{a = acceleration \ (constant)}

\boxed{d = distance \ travelled}

We know the flea attains a takeoff velocity of 1.0 m/s over a distance of 0.50 mm.

The flea starts from rest, so the initial velocity is zero. The question asks for the flea's acceleration during leg extension.

The equation we apply is:

\boxed{ \ v^2 = u^2 + 2ad \ }

  • a = acceleration (m/s²)
  • u = initial speed = 0 m/s
  • v = takeoff velocity = 1.0 m/s
  • d = displacement = 0.50 mm

First, convert 0.50 mm to meters: \boxed{0.50 \ mm = 0.50 \times 10^{-3} \ m = 5.0 \times 10^{-4} \ m}

Solution steps:

Rearrange the formula to isolate acceleration (a).

\boxed{ \ v^2 = u^2 + 2ad \ }

\boxed{ \ v^2 - u^2 = 2ad \ }

\boxed{ \ 2ad = v^2 - u^2 \ }

\boxed{ \ a = \frac{v^2 - u^2}{2d} \ }

Insert the given values into the rearranged equation.

\boxed{ \ a = \frac{(1.0)^2 - (0)^2}{2(5.0 \times 10^{-4})} \ }

\boxed{ \ a = \frac{1}{10 \times 10^{-4}} \ }

\boxed{ \ a = \frac{1}{1 \times 10^{-3}} \ }

\boxed{ \ a = 1 \times 10^{3}\ = 1,000 \ m/s^2 \ }

This yields the flea's acceleration as 1,000 m/s².

Additional resources

  1. Scientific notation: brainly.com/question/7263463
  2. Determining substance mass: brainly.com/question/4053884
  3. Conversion into cubic units: brainly.com/question/1446243

Keywords: flea, jumping, leg extension, takeoff velocity, initial speed, displacement, acceleration, constant acceleration, unit conversion

5 0
1 month ago
Read 2 more answers
The sharks are fed three times a day. During the morning feeding , 2/15 tons of fish is fed. During the afternoon feeding, the w
zzz [9080]

Answer:

1/6 ton of fish is provided during the night feeding.

Step-by-step explanation:

The fish consumed in the morning is \dfrac{2}{15} tons, and the amount of fish served in the afternoon is \dfrac{1}{15} greater than in the morning. This means

\dfrac{2}{15} +\dfrac{1}{15}

Let’s represent the quantity of fish given at night as x, and if the total fish fed throughout the day equals \dfrac{1}{2}, we have

\dfrac{2}{15}+(\dfrac{2}{15} +\dfrac{1}{15})+x=\dfrac{1}{2}

By summing the numerators on the left side, we derive:

\dfrac{5}{15} +x=\dfrac{1}{2}

and subtracting \dfrac{5}{15} from both sides allows us to isolate x:

x=\dfrac{1}{2} -\dfrac{5}{15}

Since \dfrac{5}{15} =\dfrac{1}{3}, we derive

x=\dfrac{1}{2}-\dfrac{1}{3}

The common denominator for the fractions is 6; hence, we write the equation in the form

x=\dfrac{3}{6}-\dfrac{2}{6}

and simplifying the numerators results in:

\boxed{x=\dfrac{1}{6} }

which indicates the amount of tons fed during the night feeding.

7 0
10 days ago
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