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2 years ago

What’s the area of the figure

Mathematics

2 answers:

Galina-37 [17]2 years ago

7 0

Answer:

693 m^2

Step-by-step explanation:

A = bh

A = (21)(33)

A = 693

Alenkasestr [34]2 years ago

6 0

Answer:

The Area of figure is $693m^2$

Step-by-step explanation:

The given figure is a parallelogram that has given:

Base = 21 m

Height = 33 m

Area of the parallelogram:

$Base * Height\\\\ = 21*33\\\\ = 693 m^2$

The area of figure is $693m^2$

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Write an expression to represent:<br> Nine more than the quotient of two and a number x

Taya2010 [7]

Answer:

9+(2/x)

Step-by-step explanation:

6 0

2 years ago

The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability

amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Also, important to remember that the standard deviation is the square root of the variance.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 69, \sigma = \sqrt{121} = 11, n = 31, s = \frac{11}{\sqrt{31}} = 1.97565$

What is the probability that the sample mean would be greater than 70.5 kilograms?

This is 1 subtracted by the pvalue of Z when X = 70.5. So

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

By the Central limit theorem

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

$Z = \frac{70.5 - 69}{1.97565}$

$Z = 0.76$

$Z = 0.76$ has a pvalue of 0.7764

1 - 0.7764 = 0.2236

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

8 0

2 years ago

A doctor released the results of clinical trials for a vaccine to prevent a particular disease. In these clinical trials, 200,0

CaHeK987 [17]

Answer:

1) No, there is not enough evidence to support the claim that the vaccine was effective (P-value=0.104) .

The null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0$

being the subindex 1 for the experimental group and subindex 2 for the control group.

2) Test statistic z=-1.26

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the vaccine was effective.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0$

The significance level is 0.01.

The sample 1 (experimental group), of size n1=100000 has a proportion of p1=0.0002.

$p_1=X_1/n_1=21/100000=0.0002$

The sample 2 (control group), of size n2=100000 has a proportion of p2=0.0003.

$p_2=X_2/n_2=30/100000=0.0003$

The difference between proportions is (p1-p2)=-0.0001.

$p_d=p_1-p_2=0.0002-0.0003=-0.0001$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{21+30}{100000+100000}=\dfrac{51}{200000}=0.00026$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.00026*0.99974}{100000}+\dfrac{0.00026*0.99974}{100000}}\\\\\\s_{p1-p2}=\sqrt{0+0.0000000025}=\sqrt{0.0000000051}=0.000071$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.00009-0}{0.000071}=\dfrac{-0.00009}{0.000071}=-1.26$

This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=P(z$

As the P-value (0.104) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the vaccine was effective.

6 0

2 years ago

Please help ASAP! BRAINLIEST to best/right answer (:

JulsSmile [24]

Just finished doing the math and the correct answer should be B: 5.3 Hope this helps:)

7 0

2 years ago

Read 2 more answers

In how many different ways can a jury of 12 people be randomly selected from a group of 40 people?

kupik [55]

Answer:

There are 5,586,853,480 different ways to select the jury.

Step-by-step explanation:

The order is not important.

For example, if we had sets of 2 elements

Tremaine and Tre'davious would be the same set as Tre'davious and Tremaine. So we use the combinations formula.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

In how many different ways can a jury of 12 people be randomly selected from a group of 40 people?

Here we have $n = 40, x = 12$ .

$C_{40,12} = \frac{40!}{12!(28)!} = 5,586,853,480$

There are 5,586,853,480 different ways to select the jury.

3 0

2 years ago