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MaRussiya [10]
3 years ago
15

Can anyone please help me with this?

Mathematics
1 answer:
wlad13 [49]3 years ago
8 0

1.

20 pounds of strawberries

5 persons

20 ÷ 5 = 20 pounds ÷ 5 persons = 4 pounds per person

Answer: 4 is the pounds per person, how much each person received

2.

20 pounds of strawberries

5 pounds per person

20 ÷ 5 = 20 pounds ÷ 5 pounds/person= 4 persons

Answer: 4 is the number of people we're splitting among

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Rory bought seven 3-packs of golf balls and five 20-pack of golf tees. how many golf items did he buy in all
Harlamova29_29 [7]
Hello there.

Question: <span>Rory bought seven 3-packs of golf balls and five 20-pack of golf tees. How many golf items did he buy in all? 

Answer: First you multiply the number of items in the pack by the number of packs bought.
7 x 3 = 21
5 x 20 = 100.
Add together the amount of items.
100 + 21 = 121.

In short, Your answer is Rory bought 121 golf items in total.

Hope This Helps You!
Good Luck Studying ^-^</span>
5 0
3 years ago
If the amount of vat paid by a customer while buyong a bag at 13℅ vat rate rs 143 ,find the cost of bag without vat​
Darya [45]

Step-by-step explanation:

vat %=13%

vat=143

vat = vat% of sp (i.e sp without vat)

143 = 13/ 100 of sp

143*100=13sp

14300/13 = sp

sp=4766.67

therefore the cost of bag without vat is rs.4766.67

6 0
3 years ago
The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 8.3 cm.
Gre4nikov [31]

a. Find the probability that an individual distance is greater than 214.30 cm

We find for the value of z score using the formula:

z = (x – u) / s

z = (214.30 – 205) / 8.3

z = 1.12

Since we are looking for x > 214.30 cm, we use the right tailed test to find for P at z = 1.12 from the tables:

P = 0.1314

 

b. Find the probability that the mean for 20 randomly selected distances is greater than 202.80 cm

We find for the value of z score using the formula:

z = (x – u) / s

z = (202.80 – 205) / 8.3

z = -0.265

Since we are looking for x > 202.80 cm, we use the right tailed test to find for P at z = -0.265 from the tables:

P = 0.6045

 

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

I believe this is because we are given the population standard deviation sigma rather than the sample standard deviation. So we can use the z test.

7 0
3 years ago
Chicago is home to 2.706 million people and the city sees approximately 55 million visitors each year. China has a population of
Norma-Jean [14]

China could expect a quantity of visitors of 8.512 × 10⁶ in Chicago.

<h3>How to apply direct proportions</h3>

A <em>direct</em> proportion is a <em>direct</em> relationship between two variables, which in this case is represented between the quantity of tourists in the city of Chicago and the population of the country of origin.

Let suppose the existence of direct proportionality and world has a population of 9000 million people, the quantity of Chinese visitors in the city of Chicago is determined by the following formula:

x = (1.393 × 10⁹ / 9 × 10⁹) × 55 × 10⁶

x = 8.512 × 10⁶

China could expect a quantity of visitors of 8.512 × 10⁶ in Chicago. \blacksquare

To learn more on direct proportions, we kindly invite to check this verified question: brainly.com/question/14389660

3 0
3 years ago
In a shipment of 850 widgets, 32 are found to be defective. At this rate, how many defective widgets could be expected in 22,000
Kobotan [32]
You can solve this using fractions, and say that 32/850 are defective. To find the amount of defective widgets out of 22,000, we can then multiply to make the denominator 22,000. To work out what we need to multiply by, we can just do 22,000 ÷ 850 = 25.88. Now if we multiply 32 by 25.88, we get our answer as 828 widgets.

I hope this helps!
3 0
4 years ago
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