The factors of 32 are 32, 16, 8, 4, 2, 1. The factors of 72 are 72, 36, 24, 18, 12, 9, 8, 6, 4, 3, 2, 1. The common factors of 32 and 72 are 8, 4, 2, 1, intersecting the two sets above.

plz mark brainliest :DD

4 0

3 years ago

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How many solutions does this linear system have?

gayaneshka [121]

Answer:

One solution: (2.5,0)

Step-by-step explanation:

We can use substitution for the solution. Substitute y=2x-5 to -8x-4y=-20 so that you end up with -8x-8x+20=-20. Next you want to add like terms which will be -16x+20=-20, next you want isolate x by subtracting 20 from both sides and leaves you with -16x=-40. Divide -16 on both sides to fully isolate x and will leave you with x=2.5. Now substiture in 2.5 for x in y=2x-5 to get y=2(2.5)-5 which will then lead to y=0.

4 0

3 years ago

Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 141 millimeters,

tiny-mole [99]

Answer:

Probability that the sample mean would be greater than 141.4 millimetres is 0.3594.

Step-by-step explanation:

We are given that Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 141 millimetres, and a standard deviation of 7.

A random sample of 39 steel bolts is selected.

Let $\bar X$ = sample mean diameter

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 diameter = 141 millimetres

$\sigma$ = standard deviation = 7 millimetres

n = sample of steel bolts = 39

Now, Percentage the sample mean would be greater than 141.4 millimetres is given by = P( $\bar X$ > 141.4 millimetres)

P( $\bar X$ > 141.4) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } } }$ > $\frac{141.4-141}{\frac{7}{\sqrt{39} } } }$ ) = P(Z > 0.36) = 1 - P(Z $\leq$ 0.36)

= 1 - 0.6406 = 0.3594

The above probability is calculated by looking at the value of x = 0.36 in the z table which has an area of 0.6406.

8 0

3 years ago