How do you solve this problem?

Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75.percent f

irinina [24]

Answer:

33%

Step-by-step explanation:

Assuming the weight of the mixture to be 100g**, then the weight of ryegrass in the mixture would be 30g.

Also, assume the weight mixture X used in the mixture is Xg, then the weight of mixture Y used in the mixture would be (100-X)g.

So we can now equate the parts of the ryegrass in the mixture as:

0.4X + 0.25(100-X) = 30

<=> 0.4X + 25 - 0.25X = 30

<=> 0.15X = 5

<=> X = 5/0.15 = 500/15 = 100/3

So the weight of mixture X as a percentage of the weight of the mixture

= (weight of X/weight of mixture) * 100%

= (100/3)/100 * 100%

= 33%

3 0

4 years ago

Consider the probability that greater than 94 out of 153 people will not get the flu this winter. Assume the probability that a

Hatshy [7]

Answer:

0.8212 = 82.12% probability that greater than 94 out of 153 people will not get the flu this winter.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

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

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$p = 0.65, n = 153$ . So

$\mu = E(X) = 153*0.65 = 99.45$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{153*0.65*0.35} = 5.9$

Consider the probability that greater than 94 out of 153 people will not get the flu this winter

This probability is 1 subtracted by the pvalue of Z when X = 94. So

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

$Z = \frac{94 - 99.45}{5.9}$

$Z = -0.92$

$Z = -0.92$ has a pvalue of 0.1788

1 - 0.1788 = 0.8212

0.8212 = 82.12% probability that greater than 94 out of 153 people will not get the flu this winter.

7 0

3 years ago

Choose the correct simplification of (4x − 3)(3x2 − 4x − 3).

tester [92]

$(4x - 3)(3x^2 - 4x - 3)$

$=4x\cdot \:3x^2+4x\left(-4x\right)+4x\left(-3\right)-3\cdot \:3x^2-3\left(-4x\right)-3\left(-3\right)$

$=4x\cdot \:3x^2-4x\cdot \:4x-4x\cdot \:3-3\cdot \:3x^2+3\cdot \:4x+3\cdot \:3$

$\mathrm{Simplify}\:4x\cdot \:3x^2-4x\cdot \:4x-4x\cdot \:3-3\cdot \:3x^2+3\cdot \:4x+3\cdot \:3:\quad 12x^3-25x^2+9$

$=12x^3-25x^2+9$

8 0

3 years ago

If 3t-7= 5 then 6t =

Musya8 [376]

3t=7+5=12
3t=12
t=4
6t=6×4=24
The answer is 24

3 0

3 years ago

The image is located at the bottom of the screen.

Setler79 [48]

Answer:

..... surface area = 16 km^2.

7 0

3 years ago