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nadezda [96]
3 years ago
15

Evaluate 11/12*12/121- -6/5*70/60

Mathematics
1 answer:
Oduvanchick [21]3 years ago
6 0

Answer:

82/55

Step-by-step explanation:

11/12 * 12/121 - -6/5 * 70/60 = 1/11 + 14/10 = 1/11 + 7/5 = 5/55 + 77/55 = 82/55

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The length of a rectangular garden is 4 less than twice the width find the perimeter and area of the garden
sergeinik [125]

Answer:

Let w be the width. Then the length is 2w-4. So:

w(2w-4)=96

2w²-4w-96=0

w²-2w-48=0

(w-8)(w+6)=0

w=8 or -6

Throwing out the negative value for w, we get a width of 8 yds, and a length of 12 yds. ☺☺☺☺

4 0
3 years ago
Tom Johnson worked 32 hours this week for $9.25 per hour. What is Tom’s gross pay for the week?
Dominik [7]

Answer:

296 dollars

Step-by-step explanation:

5 0
3 years ago
Solve for x and graph the solution on the number line<br> 15 &gt; 2x – 9 &gt; -23
Monica [59]

The answer is 12>x>-7

Step-by-step explanation:

it's first divided into 2 parts

1 . 15 > 2x - 9

2. 2x - 9 > -23

Then solve them individually.

let's do it,

<u>First half</u>

15 > 2x - 9

15+9 > 2x

24 > 2x

24/2 > x

12>x

<u>Second half</u>

2x-9>-23

2x>-23+9

2x>-14

x>-14/2

x>-7

<u>Joining of the two halves</u>

We have :

12>x>-7

8 0
3 years ago
An Epson inkjet printer ad advertises that the black ink cartridge will provide enough ink for an average of 245 pages. Assume t
Neko [114]

Answer:

35.2% probability that the sample mean will be 246 pages or more

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 245 \sigma = 15, n = 33, s = \frac{15}{\sqrt{33}} = 2.61

What the probability that the sample mean will be 246 pages or more?

This is 1 subtracted by the pvalue of Z when X = 246. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{246 - 245}{2.61}

Z = 0.38

Z = 0.38 has a pvalue of 0.6480.

1 - 0.6480 = 0.3520

35.2% probability that the sample mean will be 246 pages or more

4 0
3 years ago
An olympian swam the 200-meter freestyle at a speed of 1.8 meters per second. An olympic runner ran the 200-meter dash in 21.3 s
Alla [95]
It would be 20
21.3
-1.8
------
19.5 rounded would be 20
3 0
3 years ago
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