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1 year ago

En la escuela Francisco I. Madero de ciudad Delicias, llevaron a cabo el festejo para

Mathematics

1 answer:

REY [17]1 year ago

4 0

Answer:

The number of unsold cakes was 2

Step-by-step explanation:

The question in English is

In the school Francisco I. Madero of Ciudad Delicias, the celebration was held for commemorate the arrival of spring, after the parade the stalls were set up of the kermesse. The first grade group bought 8 cakes and sold 3/4 of the total.

How much of the cake was not sold?

Let

x ----> number of cakes sold

y ----> number of cakes that didn't sell

we know that

The first grade group bought 8 cakes

$x+y=8$ -----> equation A

The first grade group sold 3/4 of the total.

$x=\frac{3}{4}(x+y)$ ---> equation B

substitute equation A in equation B

$x=\frac{3}{4}(8)=6$

Find the value of y

$6+y=8$

$y=2$

therefore

The number of unsold cakes was 2

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2 years ago

Which of the following functions describes the sequence 4, –2, 1, –12, 14, . . .? A. f(1) = 4, f(n + 1) = –2f(n) for n ≥ 1 B. f(

AlekseyPX

Answer: D. $f(1) = 4,\ \ f(n+1)=\dfrac{-1}{2}f(n)$

Step-by-step explanation:

The given sequence: $4, -2,1,\dfrac{-1}{2},\dfrac14,....$

Here, first term: $f(1)=4$

Second term: $f(2)=-2$

Third term : $f(3)=\dfrac{-1}{2}$

It can be observed that it is neither increasing nor decreasing sequence but having the common ratio.

Common ratio: $r=\dfrac{f(2)}{f(1)}=\dfrac{-2}{4}=\dfrac{-1}{2}$

So, $f(n+1)=\dfrac{-1}{2}f(n)$ [as in G.P. nth term= $a_{n+1}=ar^n$ ]

Hence, correct option is D. $f(1) = 4,\ \ f(n+1)=\dfrac{-1}{2}f(n)$

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2 years ago

A process manufactures ball bearings with diameters that are normally distributed with mean 25.1 mm and standard deviation 0.08

marta [7]

Answer:

(a) The proportion of the diameters are less than 25.0 mm is 0.1056.

(b) The 10th percentile of the diameters is 24.99 mm.

(d) The proportion of the ball bearings meeting the specification is 0.8881.

Step-by-step explanation:

Let X = diameters of ball bearings.

The random variable X is normally distributed with mean, μ = 25.1 mm and standard deviation, σ = 0.08 mm.

To compute the probability of a Normally distributed random variable we need to first convert the raw scores to z-scores as follows:

z = (X - μ) ÷ σ

(a)

Compute the probability of X < 25.0 mm as follows:

P (X < 25.0) = P ((X - μ)/σ < (25.0-25.1)/0.08)

= P (Z < -1.25)

= 1 - P (Z < 1.25)

= 1 - 0.8944

= 0.1056

*Use a z-table for the probability.

Thus, the proportion of the diameters are less than 25.0 mm is 0.1056.

(b)

The 10th percentile implies that, P (X < x) = 0.10.

Compute the 10th percentile of the diameters as follows:

P (X < x) = 0.10

P ((X - μ)/σ < (x-25.1)/0.08) = 0.10

P (Z < z) = 0.10

z = -1.282

The value of x is:

z = (x - 25.1)/0.08

-1.282 = (x - 25.1)/0.08

x = 25.1 - (1.282 × 0.08)

= 24.99744

≈ 24.99

Thus, the 10th percentile of the diameters is 24.99 mm.

(c)

Compute the value of P (X < 25.2) as follows:

P (X < 25.2) = P ((X - μ)/σ < (25.2-25.1)/0.08)

= P (Z < 1.25)

= 0.8944

≈ 0.84

*Use a z-table for the probability.

Thus, the ball bearing that has a diameter of 25.2 mm is at the 84th percentile.

(d)

Compute the value of P (25.0 < X < 25.3) as follows:

P (25.0 < X < 25.3) = P ((25.0-25.1)/0.08 < (X - μ)/σ < (25.3-25.1)/0.08)

= P (-1.25 < Z < 2.50)

= P (Z < 2.50) - P (Z < -1.25)

= 0.99379 - 0.10565

= 0.88814

≈ 0.8881

*Use a z-table for the probability.

Thus, the proportion of the ball bearings meeting the specification is 0.8881.

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2 years ago

Alice and Betsy agreed to meet at 3:00PM and o shopping in one of the 6 clothing stores in their hometown.. However, they forgot

Norma-Jean [14]

They could miss each other 30 different ways. I solved this by drawing it out six different stores and making Alice and Betsy go to each one may different times. Example photo included for how I got the first five. Then I did the same thing, but switch A and B, counted that towards the total and continued the is drawing until all the possibilities were found.

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2 years ago

What is the maximum number of pages the report can have so that it can be completely stored on a USB drive that holds 4 \cdot 10

lana [24]

Answer:

2,300,000pages report.

Step-by-step explanation:

Given a USB drive with memory of 4.6*10⁶KB, in order to know the maximum number of pages a report can have so that it can be completely stored on the drive, the conversion factor must be used.

We must first understand that 1 page of a document uses up approximately 2kB of the storage.

Since 2kB = 1page

4.6*10⁶KB = x page

Cross multiply

2kB * x = 4.6*10⁶KB * 1

2x = 4.6*10⁶

x = 4.6*10⁶/2

x = 2.3*10⁶

x = 2,300,000

Hence the maximum number of pages that the report can have so that it can be completely stored on a USB drive that holds 4.6*10⁶KB is approximately 2,300,000pages report.

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2 years ago