GIIVNG BRAINLY FOR CORRECT ANSWER

In Lixue’s garden, the green pepper plants grew 5 inches in 34 month. At this rate, how many feet can they grow in one month? (L

beks73 [17]

Answer:

<h2>0.15 inches per month</h2>

Step-by-step explanation:

step one:

This problem is focused on the growth rate of a plant.

given data

the plants grow 5 inches in 34 months

and also that 1 month= 4 weeks

step two:

We can still say that

if 34 months see a growth of 5 inches

then 1 months will be X inches

cross multiply we have

X= 5/32 inches

step three:

converting this number to decimal we have

X= 0.15 inches per month

Therefore the growth rate would be

0.15 inches per month

5 0

3 years ago

Write each phrase as a variable expression:<br><br> 9 subtracted from x

Schach [20]

Answer:

9-x

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Ms. Brown predicted that 25% of the days in January would have temperatures below 45°F. How many days does she predict will have

Dmitriy789 [7]

The answer is 7.75. Since there can not be 7.75 days, you can round that to 8 and say that it's almost 8 days. Hope this helps!

8 0

3 years ago

Suppose that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in t

hram777 [196]

Answer:

a. $\mu=15$

b. $\mu=7.8586 \ and \ \mu=22.1414$

c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

Step-by-step explanation:

a.Binomial distribution is defined by the expression

$P(X=k)=C_k^n.p^k.(1-p)^{n-k}$

Let n be the number of trials, $n=100$

and p be the probability of success, $p=15\%$

The mean of a binomial distribution is the probability x sample size.

$\mu=np=100\times0.15=15$

b.Limits within which p is approximately 95%

sd of a binomial distribution is given as: $\sigma=\sqrt npq\\q=1-p$

Therefore, $\sigma=\sqrt(100\times0.015\times0.85)=3.5707$

Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.

$sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414$

Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).

c. $x=45$ is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

4 0

3 years ago

If outliers are discarded, then the retirement savings by residents of Econistan is normally distributed with a mean of $100,000

zavuch27 [327]

Answer:

$P(X>117000)=P(\frac{X-\mu}{\sigma}>\frac{117000-\mu}{\sigma})=P(Z>\frac{117000-100000}{20000})=P(z>0.85)$

And we can find this probability using the complement rule and the normal standard table or excel:

$P(z>0.85)=1-P(z$

The firgure attached illustrate the problem

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the retirement savings of a population, and for this case we know the distribution for X is given by:

$X \sim N(100000,20000)$

Where $\mu=100000$ and $\sigma=20000$

We are interested on this probability

$P(X>117000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>117000)=P(\frac{X-\mu}{\sigma}>\frac{117000-\mu}{\sigma})=P(Z>\frac{117000-100000}{20000})=P(z>0.85)$

And we can find this probability using the complement rule and the normal standard table or excel:

$P(z>0.85)=1-P(z$

The firgure attached illustrate the problem

7 0

3 years ago