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serious [3.7K]
3 years ago
11

13(y+3)=13y+39 how many solution

Mathematics
1 answer:
Jobisdone [24]3 years ago
4 0

13(y+3)=13y+39

mutiply the bracket by 13

(13)(y)= 13y

(13)(3)= 39

13y+39= 13y+39

move 13y to the right side

sign changes from +13y to -13y

13y-13y+39= 13y-13y+39

0+39= 0+39

39= 39

Answer:

all real numbers

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Which box is greater,a 1.5 lb box of raisins or a 650 gram box of raisins
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Old Navy has a sale on $15 shirts: buy 3
UNO [17]

The amount that the would be paid after 40% discount is $27

<h3>Word Problem</h3>

Given Data

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  • Discount = 40%

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Learn more about word problem here:

brainly.com/question/25693822

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7 0
3 years ago
The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than av
34kurt

Answer:

Probability that the 50 randomly selected laptops will have a mean replacement time of 3.1 years or less is 0.0092.

Yes. The probability of this data is unlikely to have occurred by chance alone.

Step-by-step explanation:

We are given that the replacement times for the model laptop of concern are normally distributed with a mean of 3.3 years and a standard deviation of 0.6 years.

He then randomly selects records on 50 laptops sold in the past and finds that the mean replacement time is 3.1 years.

<em>Let M = sample mean replacement time</em>

The z-score probability distribution for sample mean is given by;

            Z = \frac{ M-\mu}{\frac{\sigma}{\sqrt{n} } }} }  ~ N(0,1)

where, \mu = population mean replacement time = 3.3 years

            \sigma = standard deviation = 0.6 years

            n = sample of laptops = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the 50 randomly selected laptops will have a mean replacement time of 3.1 years or less is given by = P(M \leq 3.1 years)

 P(M \leq 3.1 years) = P( \frac{ M-\mu}{\frac{\sigma}{\sqrt{n} } }} } \leq \frac{ 3.1-3.3}{\frac{0.6}{\sqrt{50} } }} } ) = P(Z \leq -2.357) = 1 - P(Z \leq 2.357)

                                                           = 1 - 0.99078 = <u>0.0092</u>  or  0.92%          

<em>So, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.357 in the z table which will lie between x = 2.35 and x = 2.36 which has an area of 0.99078.</em>

Hence, the required probability is 0.0092 or 0.92%.

Now, based on the result above; <u>Yes, the computer store has been given laptops of lower than average quality</u> because the probability of this data is unlikely to have occurred by chance alone as the probability of happening the given event is very low as 0.92%.

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3 years ago
Be granted 10 points plus brainliest
BartSMP [9]

Answer:

8 square feet

Step-by-step explanation:

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3 years ago
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