All categories

3 years ago

PLS HELP I WILL GIVE BRAINLYEST 68% of what number is 137.9 m?

Mathematics

2 answers:

kherson [118]3 years ago

8 0

Answer:

968 is the answer for the question... Plz mark me as branlist answer

frutty [35]3 years ago

8 0

Answer:

238

Step-by-step explanation:

You might be interested in

Dolores spent $ 13.00 of the $20.00 in her wallet. Which decimal represents the fraction of the $20.00 Dolores spent?

Veronika [31]

Answer:

In order to find the fraction of the 20 dollars that has been spent, we have to divide the amount spent by the total amount. So 0.65 is the fraction of 20 that is spent.

Step-by-step explanation:

6 0

3 years ago

Nine less than 15 times a number is 6 write as a equation

Mila [183]

Let the number = x

You have 15 times the number so that is 15x

9 less would be subtracting 9:

15x -9

Is 6 means the equation equals 6:

The equation is: 15x-9 = 6

7 0

3 years ago

g Determine the point estimate of the population mean and margin of error for the confidence interval. Lower bound is 21, upper

nata0808 [166]

Answer: The point estimate of the population mean is 23. The margin of error for the confidence interval is 2.

Step-by-step explanation:

The confidence interval for the mean is of the form is given by :-

$\overline{x}-\text{error}\leq\mu\leq\overline{x}+\text{error}$

Given : Lower bound: $\overline{x}-\text{error}=21$

Upper bound : $\overline{x}+\text{error}= 25$

Then ,the sum of lower and upper bound will be :-

$2\overline{x}=21+25=46\\\\\Rightarrow\ \overline{x}=23$

Since $\overline{x}-\text{error}=21$

$\Rightarrow\ \text{Error}=\overline{x}-21=23-21=2$

Hence, The point estimate of the population mean is 23. The margin of error for the confidence interval is 2.

4 0

4 years ago

Consider the first five steps of the derivation of the Quadratic Formula.

lara31 [8.8K]

Answer:

Full proof below

Step-by-step explanation:

$\displaystyle ax^2+bx+c=0\\\\ax^2+bx=-c\\\\x^2+\biggr(\frac{b}{a}\biggr)x=-\frac{c}{a}\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\biggr(\frac{b}{2a}\biggr)^2=-\frac{c}{a}+\biggr(\frac{b}{2a}\biggr)^2\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=\frac{b^2-4ac}{4a^2}\\ \\x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$