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alexandr1967 [171]
2 years ago
11

A dentist wants to find out how often her patients floss their teeth. Which samples are biased? Check all that apply. all her pa

tients patients with no cavities patients younger than 18 every 20th patient every patient with braces
Mathematics
2 answers:
natta225 [31]2 years ago
8 0

Answer: the answers are options B,C,and E or

patients with no cavities

patients younger than 18

every patient with braces

Step-by-step explanation: i got them right on edge

ArbitrLikvidat [17]2 years ago
7 0

Answer:

all her patients patients with no cavities

patients younger than 18

every patient with braces

Step-by-step explanation:

when a sample is selected in a manner that some elements, in this case patients, of population have higher or lower probability of sampling then that sample is biased.

From given case, all the following are biased samples

all her patients patients with no cavities

patients younger than 18

every patient with braces

because they are non-random sample of a population in which all other elements were not equally likely to be chosen!

You might be interested in
A composition of transformations ΔKLM maps ΔK"L"M" to . Triangle K L M is rotated 270 degrees about point P to form Triangle K p
telo118 [61]

Answer:

a 270° rotation about point P

Step-by-step explanation:

Because after transformation MK and M'K 'are at 90°, that implies that MK is rotated as 90 ° clockwise around P or 270 ° counterclockwise around P.

Therefore the transformation composition that maps \triangle KLM to \triangleK"L"M",

So, for this composition, the first transformation is 270° is the 270° rotation i.e to be counterclockwise about point P.

While the transformation of the second would be down translation that goes to the right.

hence, the last option is correct

6 0
2 years ago
Read 2 more answers
A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates w
sveta [45]

Answer:

(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412  

(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318  

And the 95% confidence interval would be given (-0.1412;-0.0318).  

We are confident at 95% that the difference between the two proportions is between -0.1412 \leq p_A -p_B \leq -0.0318

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".  

p_A represent the real population proportion for business  

\hat p_A =\frac{75}{400}=0.1875 represent the estimated proportion for Business

n_A=400 is the sample size required for Business

p_B represent the real population proportion for non Business

\hat p_B =\frac{137}{500}=0.274 represent the estimated proportion for non Business

n_B=500 is the sample size required for non Business

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96  

And replacing into the confidence interval formula we got:  

(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412  

(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318  

And the 95% confidence interval would be given (-0.1412;-0.0318).  

We are confident at 95% that the difference between the two proportions is between -0.1412 \leq p_A -p_B \leq -0.0318

7 0
2 years ago
EAB and DCB are two right triangles. The figure has BED≅ BDE. Point B is the midpoint of segment AC. Prove: EAB≅ DCB
aleksklad [387]
See the attached picture. 

<span>you are given that ABCE is an isosceles trapezoid. </span>

<span>you are given that AB is parallel to EC. </span>

<span>this means that AE is congruent to BC. </span>

<span>you are given that AE and AD are congruent. </span>

<span>triangle EAD is an isosceles triangle because AE and AD are congruent. </span>

<span>this means that angle 1 is equal to angle 3. </span>

<span>since angle 1 is equal to angle 2 and angle 3 is equal to angle 1, then angle 3 is also equal to angle 2. </span>

<span>this means that AD and BC are parellel because their corresponding angles (angles 3 and 2) are equal. </span>

<span>since AB is parallel to EC and DC is part of the same line, than AB is parallel to DC. </span>

<span>you have AB parallel to DC and AD parallel to BC. </span>

<span>if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. </span>

<span>that might be able to do it,depending on whether all these statements are acceptable without proof. </span>

<span>they are either postulates or theorems that have been previously proven. </span>

<span>if not, then you need to go a little deeper and prove some of the statements that you used.. </span>

here's my diagram. 

 

<span>this is not a formal proof, but should give you some ideas about how to proceed. </span>

<span>you can also prove that angle 4 is equal to angle 2 because they are alternate interior angles of parallel lines. </span>

<span>you can also prove that angle 6 is equal to angle 5 because they are alternate interior angles of parallel lines. </span>
5 0
2 years ago
Choose all sequences of transformations that produce the same image of a given figure.
elena-14-01-66 [18.8K]

<em><u>Your answer: </u></em> a reflection across the y-axis followed by a clockwise rotation 90° about the origin. A clockwise rotation 90° about the origin followed by a reflection across the x-axis. A counter-clockwise rotation 90° about the origin followed by a reflection across the y-axis. A reflection across the x-axis followed by a counter-clockwise rotation 90° about the origin.

Hope this helps <3

Stay safe

Stay Warm

-Carrie

Ps. It would mean a lot if you marked this brainliest

5 0
2 years ago
Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro
bearhunter [10]

The question is incomplete. Here is the complete question:

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.95 probability that he will hit it. One day, Samir decides to attempt to hit  10 such targets in a row.

Assuming that Samir is equally likely to hit each of the 10 targets, what is the probability that he will miss at least one of them?

Answer:

40.13%

Step-by-step explanation:

Let 'A' be the event of not missing a target in 10 attempts.

Therefore, the complement of event 'A' is \overline A=\textrm{Missing a target at least once}

Now, Samir is equally likely to hit each of the 10 targets. Therefore, probability of hitting each target each time is same and equal to 0.95.

Now, P(A)=0.95^{10}=0.5987

We know that the sum of probability of an event and its complement is 1.

So, P(A)+P(\overline A)=1\\\\P(\overline A)=1-P(A)\\\\P(\overline A)=1-0.5987\\\\P(\overline A)=0.4013=40.13\%

Therefore, the probability of missing a target at least once in 10 attempts is 40.13%.

6 0
2 years ago
Read 2 more answers
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