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2 years ago

7

a vase has 6 flowers in it. there are 3 roses, 1/3 are peonies, and the rest are lilies. what fraction of the flowers are lilies

?

2 answers:

Readme [11.4K]2 years ago

7 0

2/3 are lilies. this is because 1/3 are peonies. this is only logical since the 2/3 would fill the remainder

hope this helps

Rudik [331]2 years ago

4 0

Hi there!

We can start off by taking out some key information from this problem.

- 6 flowers in the vase.
- 3 are roses. (3/6)
- 2 are peonies. (2/6)
- 1 is a Lily. (1/6)

By converting all of the flowers to have a common denominator we can clearly see that 1/6 of the flowers are lilies.

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As a fraction it is 466666/1000000

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cestrela7 [59]

Answer:

(a) $-\sqrt{10}y$

(b) $-\sqrt{3}y$

Step-by-step explanation:

(a) $\sqrt{10y^{2}}$ can also be written as $\sqrt{10}$ x $\sqrt{y^{2}}$

the $\sqrt{y^{2}}$ = ± y

Therefore :

$\sqrt{10y^{2}}$ = ± $\sqrt{10}y$

since it was given that y < 0 , then the factor will be $-\sqrt{10}y$

(b) $\sqrt{3y^{2}}$ can also be written as : $\sqrt{3}$ x $\sqrt{y^{2}}$

Therefore , one of the factor is $-\sqrt{3}y$

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3 years ago

He price of a shampoo, cut, and style at the hairstyling salon where you work is $18.00. You generally get a 20% tip from each c

Marianna [84]

Answer:

total earning is $64.80

Step-by-step explanation:

Price of one work = $18.00

Payment by owner = 1/4 of $18.00

= 18/4

= $4.50

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Total earning from one work = $4.50+$3.60

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2 years ago

To find the standard deviation of the diameter of wooden dowels, the manufacturer measures 19 randomly selected dowels and find

loris [4]

Answer:

Option D - $[0.12$

Step-by-step explanation:

Given : The manufacturer measures 19 randomly selected dowels and finds the standard deviation of the sample to be s=0.16.

To find : The 95% confidence interval for the population standard deviation sigma?

Solution :

Number of sample n=19

The degree of freedom is Df=n-1=19-1=18

The standard deviation of the sample is s=0.16

Applying chi-square table to find critical value,

Upper critical value of $\chi^2$ is $UC=\chi(\frac{0.05}{2},18) = 31.5264$

Lower critical value of $\chi^2$ is

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Lower limit of the 95% confidence interval for the population variance

$L=\frac{(df)\times (s^2)}{UC}$

$L=\frac{18\times (0.16^2)}{31.5264}$

$L=\frac{18\times0.0256}{31.5264}$

$L=\frac{0.4608}{31.5264}$

$L=0.0146$

Upper limit of the 95% confidence interval for the population variance

$U=\frac{(df)\times(s^2)}{LC}$

$U=\frac{18\times (0.16^2)}{8.2307}$

$U=\frac{18\times0.0256}{8.2307}$

$U=\frac{0.4608}{8.2307}$

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$[0.1208$

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2 years ago

What is the goal of your conclusion?

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Yes i think the evidence

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