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3 years ago

7

To make a specific shade of green paint, Timia mixes 2 cups of blue paint with 10 cups of yellow point. How many cups of yellow

paint should she mix with one cup of blue paint to make the same shade of green.

1 answer:

7nadin3 [17]3 years ago

7 0

Answer:

1 cup of blue paint required 5 cups of yellow paints

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Step-by-step explanation:

Given

2 cups of blue = 10 cups of yellow

Required

Determine the number of cups of yellow for 1 cup of blue paint

The given parameter shows a linear relationship between paints of blue and paints of yellow

We have:

2 cups of blue = 10 cups of yellow

Multiply both sides by 0.5

0.5* 2 cups of blue = 0.5* 10 cups of yellow

1 cup of blue = 5 cups of yellow

Hence;

<em>1 cup of blue paint required 5 cups of yellow paints</em>

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Find the sum of 2xSquared+3x-4,8-3x, and -5x squared+2

Tomtit [17]

<span>2x^2 +3x -4 + 8 - 3x -5x^2 +2
answer is

</span>-3x^2 +6

hope that helps

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3 years ago

Considering the number of questions incorrect from classmates on a quiz {10, 11, 12, 13, 13, 13, 14, 15, 16, 16, 17, 18, 18, 19,

IrinaK [193]

Answer:

According to the Empirical Rule, 68% of the data should fall between 11.98 and 18.02

Step-by-step explanation:

We are given the following data in the question:

10, 11, 12, 13, 13, 13, 14, 15, 16, 16, 17, 18, 18, 19, 20

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{225}{15} = 15$

Sum of squares of differences = 25 + 16 + 9 + 4 + 4+ 4 + 1 + 0+ 1+ 1 + 4 + 9 + 9+ 16 + 25 = 128

$S.D = \sqrt{\dfrac{128}{14}} = 3.02$

Empirical rule:

According to this rule almost all the data lies within three standard deviation of the mean for a normal distribution.
About 68% of data lies within one standard deviation of the mean.
About 95% of data lies within two standard deviations of mean.
Arround 99.7% of data lies within three standard deviation of mean.

Thus, by empirical rule,

$\mu \pm 1\sigma = 15\pm (3.02) = (11.98,18.02)$

According to the Empirical Rule, 68% of the data should fall between 11.98 and 18.02

5 0

3 years ago

Determine the intercepts of the line.<br> y = 10x - 32<br> y-intercept: (<br> z-intercept: (

Rom4ik [11]

Answer:

see below

Step-by-step explanation:

y = 10x - 32

To find the x intercept set y =0 and solve for x

0 = 10x -32

Add 32 to each side

32 = 10x

Divide by 10

32/10=10x/10

3.2 = x

To find the y intercept set x =0 and solve for t

y = 10*0 -32

y= -32

5 0

3 years ago

Read 2 more answers

Can someone help ? ASAP thanks

Ahat [919]

Answer:

2,5 then 17 20 then keep adding 15 for x and y

6 0

3 years ago

When a scientist conducted a genetics experiments with peas, one sample of offspring consisted of 943 peas, with 717 of them hav

pshichka [43]

Using the normal approximation to the binomial distribution, it is found that:

a) 0.242 = 24.2% probability of getting 717 or more peas with red flowers.

b) Since Z < 2, 717 peas with red flowers is not significantly high.

c) Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

For each pea, there are only two possible outcomes. Either they have a red flower, or they do not. The probability of a pea having a red flower is independent of any other pea, which means that the binomial distribution is used to solve this question.

Binomial distribution:

Probability of x successes on n trials, with p probability.

Normal distribution:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

If Z > 2, the result is considered <u>significantly high</u>.

If $np \geq 10$ and $n(1-p) \geq 10$ , the binomial distribution can be approximated to the normal with:

$\mu = np$

$\sigma = \sqrt{np(1-p)}$

In this problem:

943 peas, thus, $n = 943$
3/4 probability of being red, thus $p = \frac{3}{4} = 0.75$ .

Applying the approximation:

$\mu = np = 943(0.75) = 707.25$

$\sigma = \sqrt{np(1-p)} = \sqrt{943(0.75)(0.25)} = 13.297$

Item a:

Using continuity correction, this probability is $P(X \geq 717 - 0.5) = P(X \geq 716.5)$ , which is <u>1 subtracted by the p-value of Z when X = 716.5</u>.

Then:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{716.5 - 707.25}{13.297}$

$Z = 0.7$

$Z = 0.7$ has a p-value of 0.758.

1 - 0.758 = 0.242

0.242 = 24.2% probability of getting 717 or more peas with red flowers.

Item b:

Since Z < 2, 717 peas with red flowers is not significantly high.

Item c:

Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

A similar problem is given at brainly.com/question/25212369

6 0

3 years ago