A toy that was originally $35 is now marked down to $28. What percentage was the toy marked down to

Mathematics

1 answer:

Mkey [24]3 years ago

4 0

Answer: 20% off

Step-by-step explanation:

35.00 times .20 = 7.00

35.00 minus 7.00 = 28.00

You might be interested in

Plz someone help me on this too plz plz

san4es73 [151]

1. 7 and -7 are both 7 units away from 0. They are 14 units away from each other. Absolute value helps because rather than 7+(-7) being 0 absolute value makes anything positive so it becomes 7+7=14 to be able to count the units between the two.

2. 5 and -5 are both 5 units away from 0. They are 10 units away from each other. Absolute value helps because rather than 5+(-5) being 0 absolute value makes anything positive so it becomes 5+5=10 to be able to count the units between the two.

3. 2 and -2 are both 2 units away from 0. They are 4 units away from each other. Absolute value helps because rather than 2+(-2) being 0 absolute value makes anything positive so it becomes 2+2=4 to be able to count the units between the two.

Hope that helps

8 0

3 years ago

Hey fellas, giving brainliest and a lot of points to who gets this correct

Ainat [17]

Answer:

$7\frac{1}{12}$ or $\frac{85}{12}$

Step-by-step explanation:

total boxes = Jills boxes + Mikes boxes

total boxes = $5\frac{1}{3} +1\frac{3}{4}$

= $\frac{16}{3} + \frac{7}{4}$

= $\frac{64+21}{12}$

= $\frac{85}{12}$

= $7\frac{1}{12}$

3 0

3 years ago

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:

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