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2 years ago

the practice of farming which includes growing crops and raising animals to provide food and other products

Mathematics

1 answer:

Umnica [9.8K]2 years ago

4 0

Agriculture is the art and science of cultivating the soil, growing crops and raising livestock. It includes the preparation of plant and animal products for people to use and their distribution to markets. Agriculture provides most of the world's food and fabrics. Thank me later .

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What is the slope of a line that is perpendicular to 3x+2y=14

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Answer:

2/3

Step-by-step explanation:

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10 POINTS! Answer ASAP

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According to government data, 20% of employed women have never been married. If 10 employed women are selected at random, what i

Ierofanga [76]

Answer:

a) $P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

b) $P(X\leq 2) = P(X=0) + P(X=1) +P(X=2)$

$P(X=0) = (10C0) (0.2)^0 (1-0.2)^{10-0}= 0.107$

$P(X=1) = (10C1) (0.2)^1 (1-0.2)^{10-1}= 0.268$

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

And replacing we got:

$P(X\leq 2) = 0.107+0.268+0.302=0.678$

c) For this case we want this probability:

$P(X\geq 8) = P(X=8) + P(X=9) +P(X=10)$

But for this case the probability of success is p =1-0.2= 0.8

We can find the individual probabilities and we got:

$P(X=8) = (10C8) (0.8)^8 (1-0.8)^{10-8} =0.302$

$P(X=9) = (10C9) (0.8)^9 (1-0.8)^{10-9} =0.268$

$P(X=10) = (10C10) (0.8)^{10} (1-0.8)^{10-10} =0.107$

And replacing we got:

$P(X \geq 8) = 0.677$

And replacing we got:

$P(X\geq 8)=0.0000779$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Bin (n=10 ,p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Solution to the problem

Let X the random variable "number of women that have never been married" , on this case we now that the distribution of the random variable is:

$X \sim Binom(n=10, p=0.2)$

Part a

We want to find this probability:

$P(X=2)$

And using the probability mass function we got:

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

Part b

For this case we want this probability:

$P(X\leq 2) = P(X=0) + P(X=1) +P(X=2)$

We can find the individual probabilities and we got:

$P(X=0) = (10C0) (0.2)^0 (1-0.2)^{10-0}= 0.107$

$P(X=1) = (10C1) (0.2)^1 (1-0.2)^{10-1}= 0.268$

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

And replacing we got:

$P(X\leq 2) = 0.107+0.268+0.302=0.678$

Part c

For this case we want this probability:

$P(X\geq 8) = P(X=8) + P(X=9) +P(X=10)$

But for this case the probability of success is p =1-0.2= 0.8

We can find the individual probabilities and we got:

$P(X=8) = (10C8) (0.8)^8 (1-0.8)^{10-8} =0.302$

$P(X=9) = (10C9) (0.8)^9 (1-0.8)^{10-9} =0.268$

$P(X=10) = (10C10) (0.8)^{10} (1-0.8)^{10-10} =0.107$

And replacing we got:

$P(X \geq 8) = 0.677$

3 0

2 years ago

Suppose you round 43,257,529 to43,300,000.to what place value did you round the number

Arlecino [84]

100,000ths place. You are welcome

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2 years ago

A horticulturist working for a large plant nursery is conducting experiments on the growth rate of a new shrub. Based on previou

valkas [14]

Answer:

$t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771$

The degrees of freedom are given by:

$df=n-1=48-1=47$

And the p value would be:

$p_v =P(t_{(47)}$

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

Step-by-step explanation:

Information given

$\bar X=1.8$ represent the sample mean for the growth

$s=0.5$ represent the sample standard deviation

$n=48$ sample size

$\mu_o =2$ represent the value that we want to compare

$\alpha=0.01$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 2cm per week, the system of hypothesis are :

Null hypothesis: $\mu \geq 2$

Alternative hypothesis: $\mu < 2$

Since we don't know the population deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771$

The degrees of freedom are given by:

$df=n-1=48-1=47$

And the p value would be:

$p_v =P(t_{(47)}$

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

6 0

2 years ago