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3 years ago

15

There are 130 m&m's in a bag. there are half as many green as red. the number of blue is equal to the number of red and gree

n combined. there are half as many yellow as blue. there are 4 more brown than yellow. how many of each color are in the bag? what is the answer

1 answer:

AleksandrR [38]3 years ago

8 0

Let x = red
Let 0.5x = green
Let (x) + (0.5x) = 1.5x = blue
Let 0.5(1.5x) = 0.75x = yellow
Let 0.75x + 4 = brown

$(x) + (0.5x) + (1.5x) + (0.75x) + (0.75 + 4) = 130$
$x + 0.5x+ 1.5x + 0.75x + 0.75 + 4 = 130$
$4.5x + 4 = 130$
$4.5x = 126$
$x = 28$

Use this value to calculate the other required numbers.
red = 28
green = 0.5x = 0.5(28) = 14
blue = 1.5x = 1.5(28) = 42
yellow = 0.75x = 0.75(28) = 21
brown = 0.75x + 4 = 0.75(28) + 4 = 25

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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

$LC=\chi(1-\frac{0.05}{2},18) = 8.2307$

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

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

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

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

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

$U=0.0559$

So, The 95% confidence interval for the population variance is [0.0146, 0.0560]

Now, The 95% confidence interval for the population standard deviation is

$[\sqrt{0.0146}$

$[0.1208$

or $[0.12$

Therefore, Option D is correct.

The 95% confidence interval for the population standard deviation is $[0.12$

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Nicholas's sand castle is 30 cm tall and he is adding sand at a rate of 12 cm per minute. Nina's sand castle is 78 cm tall and s

Brums [2.3K]

Answer:

8 minutes

Step-by-step explanation:

Speed or Rate = Distance / time

Distance = Speed × time

Nicholas's sand castle is 30 cm tall and he is adding sand at a rate of 12 cm per minute.

Nicholas = Nina

30cm + (12cm/min)x

= 30 + 12x

Nina's sand castle is 78 cm tall and she is adding more sand to it at a rate of 6 cm per minute.

78cm + (6cm/min)x

78 + 6x

30 + 12x = 78 + 6x

12x - 6x = 78 - 30

6x = 48

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3 years ago

Can anyone help me here asapp,, I am in this question for nearly an hour

zhannawk [14.2K]

Answer:

See below

Step-by-step explanation:

Let side AB equal x. Since triangle ABC is equilateral, sides AB, BC, and Ac are all the same length, x. In any isosceles triangle(equilateral is a type of isosceles triangle) the median is the same as the altitude and angle bisector. This means we can say that AD is also a median. A median splits a side into two equal sections, so we can say BD = DC = x / 2. We are given that DC = CE, so we can also say CE = DC = x / 2. Now, we can use the pythagorean theorem to find the length of AD. So we get the equation:

AB^2 - BD^2 = AD^2

We have the values of AB and BD, so we can substitute them and solve for AD:

x^2 - (x/2)^2 = AD^2

x^2 - x^2 / 4 = AD^2

AD^2 = 3x^2 / 4

AD = x√3 / 2

DE is equal to the sum of DC and CE because of segment addition postulate, so we can say DE = DC + CE = x / 2 + x/ 2 = x. We can again use the pythagorean theorem to find the length of AE:

AD^2 + DE^2 = AE^2

(x√3 / 2)^2 + x^2 = AE^2

3x^2 / 4 + x^2 = AE^2

AE^2 = 7x^2 / 4

AE = x√7 / 2

Now, we know(from before) that AE squared is 7x^2 / 4. We can say EC squared is x^2 / 4 because EC is x / 2 and x / 2 squared is x^2 / 4. We can also notice that AE squared is 7 times EC squared because 7x^2 / 4 = 7 * x^2 / 4

Therefore, we can come to the conclusion AE^2 = 7 EC^2

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3 years ago