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3 years ago

The sum of 9 is multiplied by -2 and the answer is 90. find my number

Mathematics

1 answer:

OLga [1]3 years ago

8 0

The number is 45 because 9 times 5 is 45 times -2 is 90

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This Assignment will be used as a QUIZ GRADE!

k0ka [10]

No one is going to do all this for 5 points...

5 0

3 years ago

Read 2 more answers

Adam currently runs about 40 miles per week, and he wants to increase his weekly mileage by 30%. How many miles will Adam run pe

BigorU [14]

Answer:

52 miles/week (12 more miles per week)

Step-by-step explanation:

In order to know the miles per week with this increasement, we'll use the rule of 3, assuming that 40 miles is the 100% innitially so:

If:

40 miles --------->100%

X miles ----------> 30%

Solving for X we have the following:

X = 30 * 40 / 100 = 12 miles

So, if Adam wants to increase his weekly mileage by 30%, he needs to run 12 more miles, and that makes a total of 52 miles/week

6 0

3 years ago

In a sample of 679 new websites registered on the Internet, 42 were anonymous (i.e., they shielded their name and contact inform

Semmy [17]

Answer:

95% Confidence interval: (0.0429,0.0791)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 679

Number of anonymous websites, x = 42

$\hat{p} = \dfrac{x}{n} = \dfrac{42}{679} = 0.0618$

95% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96$

Putting the values, we get:

$0.0618\pm 1.96(\sqrt{\dfrac{0.0618(1-0.0618)}{679}}) = 0.0618\pm 0.0181\\\\=(0.0429,0.0791)$

is the required confidence interval for proportion of all new websites that were anonymous.

8 0

3 years ago

Compared to the basic savings account

kolbaska11 [484]

I have a savings accout wit four dollars in it

6 0

3 years ago

F(x)=X over x^3-2x^2+5x why will this have no zeros

Mumz [18]

If you evaluate directly this function at x=0, you'll see that you have a zero denominator.

Nevertheless, the only way for a fraction to equal zero is to have a zero numerator, i.e.

$\dfrac{x}{x^3-2x^2+5x}=0\iff x=0$

So, this function can't have zeroes, because the only point that would annihilate the numerator would annihilate the denominator as well.

Moreover, we have

$\displaystyle \lim_{x\to 0} \dfrac{x}{x^3-2x^2+5x} = \lim_{x\to 0} \dfrac{x}{x(x^2-2x+5)} = \lim_{x\to 0} \dfrac{1}{x^2-2x+5} = \dfrac{1}{5}$

So, we can't even extend with continuity this function in such a way that $f(0)=0$

5 0

3 years ago