All categories

3 years ago

In which of the following equations is the distributive property correctly applied to

Mathematics

1 answer:

kykrilka [37]3 years ago

3 0

Answer:

None, unless c was a typo

Step-by-step explanation:

Distributing gets us xy+2x=5

None of the options is that

If you meant option c as xy+2x = 5 though, then it would be correct

You might be interested in

I NEED ANSWER ASAP. BRAINLIEST

vredina [299]

Answer:

Wait I just realized, none of them are cubic? for a polynomial to be cubic it needs to be $x^{3}$

7 0

2 years ago

Plz Helppppp Which relation is a function?

gtnhenbr [62]

Option A. is a function...

here For different values of x in Domain of(x,y) ;There is a unique image y in codomain.

All others have more than one image for a single value x.

Hope it helps...

Regards;

Leukoniv/Olegion.

7 0

3 years ago

Read 2 more answers

A miner has 8 diamonds. He has 48 workers. Explain how each worker can get a piece of the diamonds.

Dmitrij [34]

Answer:

The worker can get a piece of the diamonds using 48÷8 which it is 6.

Step-by-step explanation:

I hope that it is what you're looking for. I hope this helps

6 0

2 years ago

Use the rational zero theorem to create a list of all possible rational zeroes of the function f(x) = 14x^7 - 4x^2 + 2

Bas_tet [7]

$f(x) = 14x^7 - 4x^2 + 2$

Use the rational zero theorem

In rational zero theorem, the rational zeros of the form +-p/q

where p is the factors of constant

and q is the factors of leading coefficient

$f(x) = 14x^7 - 4x^2 + 2$

In our f(x), constant is 2 and leading coefficient is 14

Factors of 2 are 1, 2

Factors of 14 are 1,2, 7, 14

Rational zeros of the form +-p/q are

$+-\frac{1,2}{1,2,7,14}$

Now we separate the factors

$+-\frac{1}{1}, +-\frac{1}{2}, +-\frac{1}{7}, +-\frac{1}{14},+-\frac{2}{1}, +-\frac{2}{2}, +-\frac{2}{7}, +-\frac{2}{14}$

$+-1, +-\frac{1}{2}, +-\frac{1}{7}, +-\frac{1}{14},+-2, +-1 , +-\frac{2}{7}, +-\frac{1}{2}$

We ignore the zeros that are repeating

$+-1, +-2, +-\frac{1}{2}, +-\frac{1}{7}, +-\frac{1}{14}, +-\frac{2}{7}$

Option A is correct

6 0

2 years ago

Read 2 more answers

The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand

Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean strength of 50 items = 74.28

$\sigma$ = population standard deviation = 2.15

n = sample of items = 50

$\mu$ = population mean tensile strength after machine was adjusted

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $74.28-1.96 \times {\frac{2.15}{\sqrt{50} } }$ , $74.28+1.96 \times {\frac{2.15}{\sqrt{50} } }$ ]

= [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.

8 0

3 years ago