Please solve this 4 3/4 * 7 2/17 A. 28 B. 32 C. 35 D. 40

It is known that 2.2< sq root 5 <2.3. Find all possible values of the expression: a sq root 5 +2 3- sq root 5 It is known

NISA [10]

Answer:

4.2 < sq. 5< 4.3

Step-by-step explanation:

since square root of 5 was +2 you just add +2 to both sides and you will get your answer, however it is different with different operations.

5 0

3 years ago

Q6) Renee buys 5kg of sweets to sell. She pays £10 for the sweets. Renee puts all the sweets into bags. She puts 250g of sweets

My name is Ann [436]

Answer:

30% percent profit

Step-by-step explanation:

100 pence in a pound.

1000 grams in a kilogram.

She bought 5000g of sweets.

She pays 10×100=1000 pence.

She put them into 5000÷250=20 bags.

She sells the bags for 65 pence each.

She sells them all for 20×65=1300 pence.

She gets a profit of 300 pence or 3 pounds.

She gets a percent profit of 30%.

3 0

3 years ago

Please help asap! Due tonight!

oksano4ka [1.4K]

Answer:

Second choice, 3

Step-by-step explanation:

a^2 + b^2 = c^2

4^2 + b^2 = 5^2

16 + b^2 = 25

b^2 = 9 (square root both sides)

b = 3

7 0

3 years ago

A random sample of 70 printers discovered that 25 of them were being used in small businesses. find the 95% limit for the popula

kvasek [131]

Answer:

The sample proportion of printers used in small business is:

$\hat{p}=\frac{25}{70}=0.3571$

The 95% confidence interval for the population proportion of printers that are used in small businesses is:

$\hat{p} \pm z_{\frac{0.05}{2}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n} }$

Where:

$z_{\frac{0.05}{2}}=1.96$ is the critical value at 0.05 significance level

$\therefore 0.3571 \pm 1.96 \sqrt{\frac{0.3571(1-0.3571)}{70} }$

$0.3571 \pm 0.1122$

$\left ( 0.3571 - 0.1122, 0.3571 + 0.1122 \right)$

$\left(0.245,0.469 \right)$

Therefore, the 95% confidence interval for the population proportion of printers that are used in small businesses is (0.245 , 0.469)

7 0

3 years ago

6, 1, -4, -9,...

kykrilka [37]

Texaschic nailed it when saying that it's arithmetic

The recursive way to write this is to say

$a_1 = 6$
$a_n = a_{n-1} - 5$

which tells us "start at 6 and each time subtract off 5"

The $a_n$ portion is the nth term while $a_{n-1}$ is the term just before the nth term

3 0

4 years ago