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vampirchik [111]

2 years ago

11

The price of a dress is reduced by 50%. When the dress still does not sell, it is reduced by 50% of the reduced price. If the

price of the dress after both reductions is $100, what was the original price?

1 answer:

Igoryamba2 years ago

5 0

Answer:

the answer is 400$

Step-by-step explanation:

You might be interested in

6= 1 - 2n + 5<br> How can I solve this equation

bulgar [2K]

Answer:

Step-by-step explanation:

Simplifying

6 = 1 + -2n + 5

Reorder the terms:

6 = 1 + 5 + -2n

Combine like terms: 1 + 5 = 6

6 = 6 + -2n

Add '-6' to each side of the equation.

6 + -6 = 6 + -6 + -2n

Combine like terms: 6 + -6 = 0

0 = 6 + -6 + -2n

Combine like terms: 6 + -6 = 0

0 = 0 + -2n

0 = -2n

Solving

0 = -2n

Solving for variable 'n'.

Move all terms containing n to the left, all other terms to the right.

Add '2n' to each side of the equation.

0 + 2n = -2n + 2n

Remove the zero:

2n = -2n + 2n

Combine like terms: -2n + 2n = 0

2n = 0

Divide each side by '2'.

n = 0

Simplifying

n = 0

8 0

3 years ago

Is a triangle with 3m 2m 4m a right triangle

lys-0071 [83]

Answer:

Step-by-step explanation:

No it’s not because 4+9 = 16

And 13 < 16 when it has to be 16=16 therefore it is anything but a right triangle

4 0

2 years ago

What is a ratio equivalent to 3:2? (not 6:1)

Akimi4 [234]

Answer:

9:6, 12:8, and 15:10 are all ratios equivalent to 3:2

Step-by-step explanation:

To find an equal ratio, you can either multiply or divide each term in the ratio by the same number (but not zero).

4 0

3 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So $n = 6, s = \frac{106}{\sqrt{6}} = 43.27$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 523}{43.27}$

$Z = -2.61$

$Z = -2.61$ has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

6 0

3 years ago

P and Q are two points on the line x - y + 1 = 0 and are at distance 5 from the origin. Find the area of the triangle OPQ.

Alecsey [184]

Answer:

P and Q are two points on the line x-y+1=0 and are at a distant of 5 units from the origin. Find the area of triangle POQ.

Step-by-step explanation:

P and Q are the intersection points of

x-y+1 = 0 and the circle x^2 + y^2 = 25

sub y = x+1 into the circle

x^2 + (x+1)^2 = 25

x^2 + x^2 + 2x + 1 - 25 = 0

x^2 + x - 12 = 0

(x+4)(x-3) = 0

x = 3 or x = -4

y = 4 or y = -3

so P(3,4) and Q(-4,3) are our two points

Height of triangle.

h = |0 - 0 + 1|/√2 = 1/√2

PQ = √( (-7)^2 + 1^2) = √50 = 5√2

area POQ = (1/2)(1/√2)(5√2) = 5/2 square units

hope this helped

7 0

2 years ago