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Charra [1.4K]
3 years ago
10

Please find the value of x please

Mathematics
1 answer:
marin [14]3 years ago
5 0
Answer is 7

Step by step explanation

3x+159=180
3x=180-159=21
x=7
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Identify the factors of 13m-2n
scZoUnD [109]
The factors of 13m are 13 and m and it goes the same with 2n hope that helps.

3 0
4 years ago
Read 2 more answers
What is the scientific notation of 0.00165?
Ierofanga [76]

Answer:

The answer is 1.65*10^{-3}

Step-by-step explanation:

when writing scientific notation, there must be one number to the left of the decimal point.

To achieve this in this problem, you need to move the decimal point 3 times to the right.

because you are moving the point to the right, the exponent will be negative.

*rule of thumb: moving the decimal to the left will give you a positive exponent, moving to the right will give you a negative exponent*

therefore you answer should be written as:

1.65*10^{-3}

hope this helps :)

6 0
3 years ago
The shape of the distribution of the time required to get an oil change at a 15-minute oil-change facility is unknown. However,
horsena [70]

Answer:

The mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.

Step-by-step explanation:

We are given that records indicate that the mean time is 16.2 minutes, and the standard deviation is 3.4 minutes.

On a typical​ Saturday, the​ oil-change facility will perform 35 oil changes between 10 A.M. and 12 P.M. Assuming that data follows normal distribution.

Let \bar X = <u><em>sample mean time</em></u>

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

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

where, \mu = population mean time = 16.2 minutes

            \sigma = standard deviation = 3.4 minutes

            n = sample size = 35 oil changes

<u>Now, the mean oil-change time that would there be a 10% chance of being at or below is given by;</u>

<u></u>

          P(X \leq x) = 0.10          {where x is required mean oil-change time}

          P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{x-16.2}{\frac{3.4}{\sqrt{35} } } ) = 0.10

          P(Z \leq \frac{x-16.2}{\frac{3.4}{\sqrt{35} } } ) = 0.10

Now, in the z table the critical value of X which represents the below 10% of the probability area is given as -1.282, that means;

                  \frac{x-16.2}{\frac{3.4}{\sqrt{35} } }  =  -1.282

               x - 16.2  =  -1.282 \times {\frac{3.4}{\sqrt{35} } }

                         x  =  16.2 - 0.74 = <u>15.46 minutes</u>

Hence, the mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.

8 0
3 years ago
-1/2 divided by 1 3/4 as a fraction in simplest form
Helen [10]

\Large\textsf{\textbf{Answer\;:}}}

\large\textsf{refer\:to\:the\:"calculations"\:section\:of\:this\:ans}

\Large\textsf{\textbf{Calculations\::}}

-\displaystyle\frac{1}{2} :1\frac{3}{4}

First , convert the mixed number (the fraction we are dividing by in the given maths problem) into an improper fraction .

Multiply the whole part (1) times the denominator (4) and add the numerator . The result is the numerator of our fraction , and the denominator stays the same .

So now our maths problem looks as follows :

-\displaystyle\frac{1}{2} :\frac{7}{4}

Now turn the fraction you are dividing by over :

-\displaystyle\frac{1}{2} *\frac{4}{7}

Now multiply the numerator of the first fraction times the numerator of the second fraction :

4

Perform the same operation with the denominators :

-4/14

So the result is :

-\displaystyle\frac{4}{14}

Since the above fraction is not in it's simplest form , we divide the top and bottom by 2 (a common factor of both 4 and 14) :

-\displaystyle\frac{2}{7}

\footnotesize\textsf{hope\:helpful~}

8 0
3 years ago
after a 7% increase in salary, laurie makes $1647.80 per month how much did she earn per month before the increase
enyata [817]

1647.80 - ( \frac{7}{100} ) \\  = 115.346 \\  \\ so \: before  \: the \: increase \:  price \:  \:  \:  \\ 1647.80 - 115.346 \\  = 1532.454
5 0
3 years ago
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