What factors would best account for the rise in population in 1850

1)The price of a sweater was slashed from #960 to $816 by a shopkeeper in the winter season. Find the rate of discount given by

Annette [7]

Answer:

The discount% is 15%

Step-by-step explanation:

Given : The price of a sweater was slashed from Rs 960 to Rs 816 by a shopkeeper in the winter season.

To find : The rate of discount given by him?

Solution :

According to question,

The market price of the sweater is Rs.960

The selling price of the sweater is Rs.816

The discount on the sweater is given bY

Discount % is given by

Therefore, The discount% is 15%.

8 0

3 years ago

I need help on #4 ASAP!!!!

kodGreya [7K]

Hi there.

A triangle's interior angles must always add up to 180 degrees. Since we already have one measurement, 56, we can set up an equation to solve for the missing angles.

(2x + 4) + 56 + x= 180; solve for x.
Subtract 56 from both sides.
(2x + 4) + x = 124;
Combine like-terms (x).
3x + 4 = 124;
Subtract 4 from both sides.
3x = 120
Divide both sides by 3 to solve for x.
x = 40.

Now, we need to substitute x with 40 in each of our angles to determine their measurements.
2x + 4; x = 40.
2(40) + 4 = 80 + 4 = 84;
One measurement is 84 degrees.

x = 40 is another measurement on its own.

Our measurements are:
56, 84, and 40.

Your corresponding answer choice is H.) 56, 84, 40.

I hope this helps!

7 0

2 years ago

What is value of x? Please help! Will reward

Andre45 [30]

X = 50
.................

8 0

3 years ago

For the month of November in a certain city, 88% of the days are cloudy. Also in the month of November in the same city, 54%

Gnoma [55]

Answer:

There is a 61.36% probability that a randomly selected day in November will be foggy if it is cloudy.

Step-by-step explanation:

We have these following probabilities:

An 88% probability that the day is cloudy.

An 54% probability that the day is both foggy and cloudy.

What is the probability that a randomly selected day in November will be foggy if it is cloudy?

This is the percentage of days that are cloudy and foggy divided by those that are cloudy. So:

$P = \frac{0.54}{0.88} = 0.6136$

There is a 61.36% probability that a randomly selected day in November will be foggy if it is cloudy.

4 0

3 years ago

Question 8 of 10

Andreyy89

Median and IQR are the most appropriate measures of center and spread for this data set.

<h3>Why are Median and IQR the most appropriate?</h3>

Among the 3 central tendencies that includes the mean, median and mode; the median is the better measure because of the followings:

Mean is affected by extreme values
Mean is not correct if more outliers are present
Mean may not represent the nature of the data whether skewed right or left.

Also, the median as the middle entry is not affected by extreme items or outliers, so the median is better than mean,

Furthermore, for the measure of spread, the IQR is better since extreme items will show higher std deviation and also some outliers mislead.

Therefore, Option B is correct.