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Reptile [31]
3 years ago
10

The battery standby duration (in hours) of a new model of cell phone is known to be normally distributed. Ten pieces of such new

model of cell phone supplied from the manufacturer are randomly chosen and the actual standby durations are recorded as below:
48.2 47.8 45.6 47.2 49.3
51.2 44.2 45.4 49.2 43.6

The manufacturer claimed that this new model of cell phone has the mean battery standby duration of longer than 46.5 hours. Test at 1% significance level if this claim is true.
Mathematics
1 answer:
emmainna [20.7K]3 years ago
8 0

x = number of hours

want to find probability (P) x >= 13

x is N(14,1) transform to N(0,1) using z = (x - mean) / standard deviation so can look up probability using standard normal probability table.

P(x >= 13) = P( z > (13 - 14)/1) = P(z > -1) = 1 - P(z < -1) = 1 - 0.1587 = 0.8413

To convert that to percentage, multiply 100, to get 84.13%

Please mark me brainliest

You might be interested in
Rectangle ABCD is dilated to create rectangle A'B'C'D' . The width of rectangle ABCD is 12 feet. The width of rectangle A'B'C'D'
Ierofanga [76]

The area of rectangle A'B'C'D' is 32 square feet

Step-by-step explanation:

A dilation is a transformation that produces an image that is the

same shape as the original, but is a different size

A dilation stretches or shrinks the original figure

The figure and its image after dilation are similar

If two rectangles are similar, then

  • \frac{w_{1}}{w_{2}}=\frac{l_{1}}{l_{2}} = constant ratio
  • \frac{P_{1}}{P_{2}}=\frac{w_{1}}{w_{2}}
  • \frac{A_{1}}{A_{2}}=(\frac{w_{1}}{w_{2}})^{2}

∵ Rectangle ABCD is dilated to create rectangle A'B'C'D'

∴ Rectangle ABCD is similar to rectangle A'B'C'D'

∵ The width of rectangle ABCD is 12 feet

∵ The width of rectangle A'B'C'D' is 8 feet

∴ \frac{w_{1}}{w_{2}}=\frac{12}{8}

- Simplify the ratio to its lowest term by divide up and down by 4

∴ \frac{w_{1}}{w_{2}}=\frac{3}{2}

∴ The scale factor of dilation is \frac{3}{2}

Let us use the rule of the similar rectangles above

∵ The area of rectangle ABCD is 72 square feet

∵ \frac{A_{1}}{A_{2}}=(\frac{w_{1}}{w_{2}})^{2}

∴ \frac{72}{A_{2}}=(\frac{3}{2})^{2}

∴ \frac{72}{A_{2}}=\frac{9}{4}

- By using cross multiplication

∴ A_{2} × 9 = 72 × 4

∴ 9 A_{2} = 288

- Divide both sides by 9

∴ A_{2} = 32 feet²

∴ The area of rectangle A'B'C'D' = 32 feet²

The area of rectangle A'B'C'D' is 32 square feet

Learn more:

You can learn more about area of shapes in brainly.com/question/6530759

#LearnwithBrainly

3 0
3 years ago
Read 2 more answers
Which one ????????????
gregori [183]
B is your answer because if u look at it the
5 0
3 years ago
HELLO I FORGOT THIS ONE PLEASE HELP
kow [346]

Answer:

No problem it's just C

4 0
3 years ago
Read 2 more answers
Hola, alguien me puede resolver esta operación? a. Múltiplo de 5 y de 6 mayor que 95.000 ( _ 8 9 1 _ ) b. Divisible por 3, 4 y 5
bezimeni [28]

Answer:

a) 98,910

b) 4,620

Step-by-step explanation:

Recordar que un numero N es multiplo de A si N es divisible por A.

Entonces, decir:

Buscar N, tal que sea multiplo de A

es lo mismo que decir:

Buscar N, tal que sea divisible por A.

a) Queremos un múltiplo de 5 y de 6 mayor que 95.000

( _ 8 9 1 _ )

Sabemos que los múltiplos de 5 terminan en 0 o 5.

Y un número de 5 dígitos mayor que 95,000 requiere que el quinto digito en este caso sea 9, entonces tendremos:

98,91_

Para encontrar el dígito de las unidades, podemos probar remplazarlo por 0 y por 5, y luego dividirlo por 6.

Si el resultado es un número entero, entonces el número será un múltiplo de 6.

Probemos con 0, vamos a tener el número:

98,910

dividiéndolo por 6 obtenemos:

98,910/6 = 16,485

Entonces:

98,910 es más grande que 95,000

es múltiplo de 5 (porque termina en cero)

Y es múltiplo de 6, como acabamos de comprobar.

b) Ahora queremos un número divisible por 3, 4 y 5, que sea menor que 4,675

4,6__  

De vuelta, como tiene que ser divisible por 5, sabemos que termina en 5 o 0.

Y como también tiene que ser divisible por 4, sabemos que tiene que terminar si o si en número par, entonces podemos concluir que solo puede terminar en 0.

Así, sabemos que el número va a ser algo como:

4,6_0

Ahora podemos probar con el dígito de las decenas, de tal forma que el número sea divisible por 3, 4 y 5.

Recordar que este número tiene que ser menor que 4,675

Entonces el número faltante puede ser:

0, 1, 2, 3, 4, 5, 6, 7

Ahora también podemos recordar que un número es solo divisible por 3 si la suma de todos sus dígitos es divisible por 3.

Entonces tenemos que tener que:

4 + 6 + x + 0 = múltiplo de 3

10 + x = múltiplo de 3

Los posibles valores de x van a ser:

x = 2   (porque 12 es múltiplo de 3)

x = 5   (porque 15 es múltiplo de 3)

Ok, ahora sabemos que el número restante solo puede ser 2 o 3.

Entonces tenemos que remplazar esos valores en el número y ver si este es divisible por 4 o no.

Comenzamos con el 2.

Tendremos el número:

4,620

Probamos dividiendo esto por 4:

4,620/4 = 1,155

El resultado es entero, entonces 4,620 es divisible por 4 (y por como lo creamos, también sabemos que es divisible por 3 y por 5, y además es menor que 4,675, como queríamos)

5 0
2 years ago
Help plz with a math problem
sergejj [24]

Answer:

Profit 527

Profit percent 0.085

8 0
3 years ago
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