How to Solve This Question,I don't understand it

Pooja's plant began sprouting 222 days before Pooja bought it, and she had it for 989898 days until it died. At its tallest, the

MakcuM [25]

Answer:

Step-by-step explanation:

Given:

function h(t) models the height of Pooja's plant (in cm) where

t is the number of days after she bought it.

Now we have to find about which number type is more appropriate for the domain of h.

That means what values can be taken by the variable "t".

As per the questions, t represents number of days not the hours so it will not be in decimal or fraction value. It can only use integer values for the number of days. like 1, 2, 3 ...,n

Now as the time is counted after she bought the plant then number of days will be positive.

Hence answer for the type of domain can be positive integers or you can say integers greater than or equal to 0.

5 0

3 years ago

Read 2 more answers

each of the 20 balls is tossed independently and at random into one of the 5 bins. let p be the probability that some bin ends u

amm1812

if p is the probability that some bin ends up with 3 balls and q is the probability that every bin ends up with 4 balls. pq is 16.

First, let us label the bins with 1,2,3,4,5.

Applying multinomial distribution with parameters n=20 and p1=p2=p3=p4=p5=15 we find that probability that bin1 ends up with 3, bin2 with 5 and bin3, bin4 and bin5 with 4 balls equals:

5−2020!3!5!4!4!4!

But of course, there are more possibilities for the same division (3,5,4,4,4) and to get the probability that one of the bins contains 3, another 5, et cetera we must multiply with the number of quintuples that has one 3, one 5, and three 4's. This leads to the following:

p=20×5−2020!3!5!4!4!4!

In a similar way we find:

q=1×5−2020!4!4!4!4!4!

So:

pq=20×4!4!4!4!4!3!5!4!4!4!=20×45=16

thus, pq = 16.

To learn more about Probability visit: brainly.com/question/29508225

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7 0

1 year ago

If (6y − 11)(6y 11) = ay^2 − b, what is the value of a? the value of a is ____________.

Triss [41]

(6y - 11)(6y + 11) = ay² - b |use (a - b)(a + b) = a² - b²

(6y)² - 11² =ay² - b

36y² - 121 = ay² - b |add b to both sides

ay² = 36y² - 121 + b |divide both sides by y² ≠ 0

a = (36y² - 121 + b)/y²

$\boxed{a=\dfrac{36y^2-121+b}{y^2}}$

7 0

3 years ago

Men are believed to have a slightly longer average length of short hospital stays than women; 5.2 days versus 4.5 days respectiv

Georgia [21]

Answer:

Since our calculated value is lower than our critical value, $z_{calc}=2.02$ , we have enough evidence to FAIL to reject the null hypothesis at 1% of singificance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

Step-by-step explanation:

1) Data given and notation

$\bar X_{1}=5.2$ represent the mean for group men

$\bar X_{2}=4.5$ represent the mean for group women

Assuming these values for the remaining data:

$\sigma_{1}=1.2$ represent the population standard deviation for the sample men

$\sigma_{2}=1.5$ represent the population standard deviation for the sample women

$n_{1}=32$ sample size for the group men

$n_{2}=30$ sample size for the group women

z would represent the statistic (variable of interest)

$p_v$ represent the p value

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for men it's higher than the mean for women, the system of hypothesis would be:

H0: $\mu_{1} \leq \mu_{2}$

H1: $\mu_{1} > \mu_{2}$

If we analyze the size for the samples both are higher than 30, and we know the population deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

We have all in order to replace in formula (1) like this:

$z=\frac{5.2-4.5}{\sqrt{\frac{1.2^2}{32}+\frac{1.5^2}{30}}}=2.02$

4) Find the critical value

In order to find the critical value we need to take in count that we are conducting a right tailed test, so we are looking on the normal standard distribution a value that accumulats 0.01 of the area on the right and 0.99 of the area on the left. We can us excel or a table to find it, for example the code in Excel is:

"=NORM.INV(1-0.01,0,1)", and we got $z_{critical}=2.33$

5) Statistical decision

Since our calculated value is lower than our critical value, $z_{calc}=2.02$ , we have enough evidence to FAIL to reject the null hypothesis at 1% of significance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

6 0

3 years ago

Find X in the given figures.

Contact [7]

The correct answer is 6

7 0

3 years ago