A gym is offering a trial membership for 3 months by offering a $50

Describe a way to make the arithmetic problem 45% x 680 mentally

lord [1]

Answer:

306.

Step-by-step explanation:

Firstly, we are aware 100% of 680 is obviously 680,

Next, we know half of a 100% is 50% therefore half of 680 is 340.

Then, we know 45% is 5% less than 50%

To find 5% of a number mentally, divide it by 10 as 100÷5=20

680÷20=34

Lastly 50% subtracted by 5%: 340-34=306.

Now we know, 45% of 680 is 306.

<em>Hope this helps.</em>

<em>Hope this helps.Please mark BRAINLIEST.</em>

5 0

3 years ago

You estimate that there are 45 marbles in a jar. The actual amount is 82 marbles. Find the percent error. Round to the nearest t

Nataly_w [17]

Answer:

45.1

Step-by-step explanation:

The original amount is 82

But the wrong amount is 45

So we will have to calculate the percentage for the error

82-45

37

So %error= 37/82 ×100

=3700/82

=45.121~45.1( to nearest tenth)

So the final answer is 45.1%

6 0

4 years ago

Question:

Andrew [12]

Answer:

(0, -9)

Step-by-step explanation:

The y intercept is the y value when x =0

(0, -9)

4 0

3 years ago

There are 360 boys in green valley elementary school elementary school there are 180 there are 180 more boys then girls

HACTEHA [7]

Your answer to this problem is 180 girls in green valley elementary school.

8 0

3 years ago

Read 2 more answers

Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the

ziro4ka [17]

Answer:

a) $\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=9-1=8$

The Confidence interval is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and the critical values are:

$\chi^2_{\alpha/2}=20.09$

$\chi^2_{1- \alpha/2}=1.65$

And replacing into the formula for the interval we got:

$\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

4 0

4 years ago