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3 years ago

7

Half of Robert's piece of wire is equal to ⅔ of Maria's wire. The total length of their wires is 12 feet. How much longer is Rob

ert's wire than Maria's?

1 answer:

____ [38]3 years ago

8 0

Answer:

Robert's piece of wire is 5 feet long

and Maria's piece of wires is 7 feet long

Step-by-step explanation:

Suppose x and y are Robert's and Maria's piece of wires respectively.

We can write following equation using this statement "Half of Robert's piece of wire is equal to ⅔ of Maria's wire".

$\frac{1}{2} x = \frac{2}{3} y$ --------------(1)

The total length of their wires is 12 feet.

$x+y=12$ ------------(2)

=> $x=12-y$

Substitute x=12-y in equation (1)

$\frac {1}{2} (12-y) = \frac{2}{3} y$

$6-\frac{1}{2} y = \frac {2}{3} y$

$6 = \frac {2} {3} y + \frac {1} {2} y$

$y=7$

Put y=7 in equation (2)

$x+7 = 12 \\x =5$

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3 years ago

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The mean amount purchased by a typical customer at Churchill's Grocery Store is $23.50 with a standard deviation of $5.00. Assum

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Answer:

a. $\mathtt{P(X \geq 25) =0.0170}$ ( to four decimal places)

b. $P(22.5$ ( to four decimal places )

c. The limits will be between the interval of ( 22.33,24.67 )

Step-by-step explanation:

Given that :

mean = 23.50

standard deviation = 5.00

sample size = 50

The objective is to calculate the following:

(a) What is the likelihood the sample mean is at least $25.00?

Let X be the random variable, the probability that the sample mean is at least 25.00 is:

$P(X \geq 25) = 1 - P(\dfrac{X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{25- 23.50}{ \dfrac{5}{\sqrt{ 50}} })$

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$P(X \geq 25) = 1 - P(Z< \dfrac{1.5 \times 7.071}{ {5}})$

$P(X \geq 25) = 1 - P(Z< 2.1213)$

$P(X \geq 25) = 1 - P(Z< 2.12)$ to two decimal places

From the normal tables :

$P(X \geq 25) = 1 - 0.9830$

$\mathtt{P(X \geq 25) =0.0170}$ ( to four decimal places)

(b) What is the likelihood the sample mean is greater than $22.50 but less than $25.00?

$P(22.5$

$P(22.5$

$P(22.5$

$P(22.5$

$P(22.5$ to four decimal places

(c) Within what limits will 90 percent of the sample means occur?

At 90 % confidence interval, level of significance = 1 - 0.90 = 0.10

The critical value for the $z_{\alpha/2} = 0.05$ = 1.65

Standard Error = $\dfrac{\sigma}{\sqrt{n}}$

Standard Error = $\dfrac{5}{\sqrt{50}}$

Standard Error = 0.7071

Therefore, at 90 percent of the sample means, the limits will be between the intervals of : $(\mu \pm z_{\alpha/2} \times S.E)$

Lower limit = ( 23.5 - (1.65×0.707) )

Lower limit = ( 23.5 - 1.16655 )

Lower limit = 22.33345

Lower limit = 22.33 (to two decimal places).

Upper Limit = ( 23.5 + (1.65*0.707) )

Upper Limit = ( 23.5 + 1.16655 )

Upper Limit = 24.66655

Upper Limit = 24.67

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3 years ago

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The slope of f(x) is steeper than the slope of g(x).

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