There are 90 students in 5th Grade. 10% of them were absent today. How many students were present .

A fence must be built to enclose a rectangular area of 45 comma 000 ftsquared. Fencing material costs $ 3 per foot for the two s

Tatiana [17]

Answer:

150 feet by 300 feet.

Step-by-step explanation:

The fence is to enclose a rectangular area of 45,000 ft squared.

If the dimensions of the rectangle are x and y

Area of a rectangle = xy

xy=45000
$x=\frac{45000}{y}$

Perimeter of the Rectangle =2x+2y

Fencing material costs $ 3 per foot for the two sides facing north and south and $6 per foot for the other two sides.

Cost of Fencing, C=$(6*2x+3*2y)=$(12x+6y)

Substitute $x=\frac{45000}{y}$ into the Cost to get C(y)

C=12x+6y

$C(y)=12(\frac{45000}{y})+6y\\C(y)=\frac{540000+6y^2}{y}$

The value at which the cost is least expensive is at the minimum point of C(y), when the derivative is zero.

$C^{'}(y)=\dfrac{6y^2-540000}{y^2}$

$\dfrac{6y^2-540000}{y^2}=0\\6y^2-540000=0\\6y^2=540000\\y^2=\frac{540000}{6} =90000\\y=\sqrt{90000}=300$

Recall,

$x=\frac{45000}{y}=\frac{45000}{300}=150$

Since x=150, y=300

The dimensions that will be least expensive to build is 150 feet by 300 feet.

6 0

3 years ago

Clarinex is a drug used to treat asthma. In clinical tests of this drug, 1655 patients were treated with 5- mg doses of Clarinex

bagirrra123 [75]

Answer:

$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$

$p_v =P(z>3.363)=0.00039$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

Step-by-step explanation:

Data given and notation

n=1655 represent the random sample taken

$\hat p=0.021$ estimated proportion of interest

$p_o=0.012$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportions is higher than 0.012.:

Null hypothesis: $p \leq 0.012$

Alternative hypothesis: $p > 0.012$

When we conduct a proportion test we need to use the z statisitic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>3.363)=0.00039$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

3 0

3 years ago

Please help me find the answer.

Sati [7]

Answer:

a = 16 and b = 2

Step-by-step explanation:

3 0

2 years ago

(2/3) To the power of 4

Serga [27]

Answer:

(2/3)^4=16/81

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Assuming boys and girls are equally likely, find the probability of a couple having a baby girl when their sixth child is born

Ivanshal [37]

Answer: The required probability of having 6th girl is 0.5.

Step-by-step explanation: Given that boys and girls are equally likely.

We are to find the probability of a couple having a baby girl when their sixth child is born, given that the first five children were all girls.

Since the events of having a boy and a girl are independent of each other, so

the probability of having 6th girl dose not depend on the birth of the first five girls.

We know that there are only two possible cases (either a boy or girl will born).

So, sample space, S = {G, B} and the event E of having a girl is, E = {G}.

That is, n(S) = 2 and n(E) = 1.

Therefore, the probability of event E is given by

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{1}{2}=0.5.$

Thus, the required probability of having 6th girl is 0.5.

3 0

3 years ago