It pumps blood away from the heart into the aorta

Find the unknown angle measures.

adell [148]

Answer:

The value of x is 47°

and the value of y is 100°

7 0

3 years ago

20 points and brainlist if correct!

weeeeeb [17]

Answer:

The Commutative property states that numbers can be multiplied in any order

Step-by-step explanation:

look at the equation 4x2.36x.25. If you write the equation differently as 4x.25x2.36, the equation becomes easier to solve. Why? As you can see 4 and .25 (aka 1/4) 4 and 1?4 are reciprocals and anything multiplied by the reciprocal is 1. so 1x 2.36 is 2.36. The Communicative Property just helps to see those patterns and connections

hope this helps ;)

6 0

2 years ago

An article presents measures of penetration resistance for a certain fine-grained soil. Fifteen measurements, expressed as a mul

balu736 [363]

Answer:

$2.64-2.14\frac{1.32}{\sqrt{15}}=1.911$

$2.64+2.14\frac{1.32}{\sqrt{15}}=3.369$

So on this case the 95% confidence interval would be given by (1.911;3.37)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X=2.64$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=1.32 represent the sample standard deviation

n=15 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=15-1=14$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,14)".And we see that $t_{\alpha/2}=2.14$

Now we have everything in order to replace into formula (1):

$2.64-2.14\frac{1.32}{\sqrt{15}}=1.911$

$2.64+2.14\frac{1.32}{\sqrt{15}}=3.369$

So on this case the 95% confidence interval would be given by (1.911;3.37)

6 0

3 years ago

Multiply.<br><br> 8.55 × 12.61<br><br> Enter your answer in the box.

ankoles [38]

Answer:

107.8155

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A population of values has a normal distribution with μ = 149.8 μ=149.8 and σ = 68.2 σ=68.2. You intend to draw a random sample

Len [333]

Answer:

There is a 49.20% probability that a single randomly selected value is less than 148.3.

There is a 38.21% probability that a sample of size n = 186 n=186 is randomly selected with a mean less than 148.3.

Step-by-step explanation:

Problems of normally distributed samples can be 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.

In this problem, we have that:

A population of values has a normal distribution with $\mu = 149.8$ and $σ=68.2$ .

Find the probability that a single randomly selected value is less than 148.3

This is the pvalue of Z when $X = 148.3$ .

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

$Z = \frac{148.3 - 149.8}{68.2}$

$Z = -0.02$

$Z = -0.02$ has a pvalue of 0.4920

There is a 49.20% probability that a single randomly selected value is less than 148.3.

Find the probability that a sample of size n = 186 n=186 is randomly selected with a mean less than 148.3.

We want to find the mean of the sample, so we have to find the standard deviation of the population. That is

$s = \frac{68.2}{\sqrt{186}} = 5$

Now, we have to find the pvalue of Z when $X = 148.3$ .

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

$Z = \frac{148.3 - 149.8}{5}$

$Z = -0.3$

$Z = -0.3$ has a pvalue of 0.3821

There is a 38.21% probability that a sample of size n = 186 n=186 is randomly selected with a mean less than 148.3.

8 0

3 years ago