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3 years ago

If the bases of an isosceles trapezoid have lengths of 13 and 26 what is the length of the median

Mathematics

2 answers:

alukav5142 [94]3 years ago

6 0

Answer: the answer is C . 19.5 units - APEX :D

Reika [66]3 years ago

5 0

Answer:

Answer is 19.5 units

Step-by-step explanation:

Given that there is a trapezoid.

Its bases are 13 and 26 respectively.

Non parallel sides are congruent

To find the median that is the length of line connecting the midpoints of the non parallel sides

If we draw any one diagonal we see that median is the line connecting the mid point of two triangles with the parallel sides as bases

Hence length of median= 1/2 (base 1+base 2)

(By triangle mid point theorem)

= 1/2 (13+26)

=19.5 units

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-8.5÷3.4 how to solve this step by step

sertanlavr [38]

You have to move both decimals over and then divide. So the answer is 25.

3 0

3 years ago

Please help! In circle Y, what is m? 82° 100° 106° 118°

ss7ja [257]

Answer:

The correct option is: 82°

Step-by-step explanation:

In the given diagram, two chords $RT$ and $SU$ are intersecting.

According to the Angle of intersecting chord theorem, "If two chords intersect inside a circle, then the measure of the angle formed is one half the sum of the measure of the arcs intercepted by the angle and its vertical angle."

That means here......

$94\°=\frac{1}{2}(m\widehat{RS}+m\widehat{TU})\\ \\ 94\°=\frac{1}{2}(106\°+m\widehat{TU})\\ \\ 2(94\°)=106\°+m\widehat{TU}\\ \\ 188\°=106\°+m\widehat{TU}\\ \\ m\widehat{TU}=188\°-106\°=82\°$

So, the measure of arc $TU$ is 82°

5 0

3 years ago

When the sample size and the sample proportion remain the same, a 90 percent confidence interval for a population proportion p w

deff fn [24]

Answer:

A 90% confidence interval for p will be narrower than the 99% confidence interval.

Step-by-step explanation:

The formula to compute the (1 - α) % confidence interval for a population proportion is:

$CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Here $\hat p$ is the sample proportion.

The margin of error of the confidence interval is:

$MOE= z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The MOE is dependent on:

Confidence level
Standard deviation
Sample size

The MOE is directly related to the confidence level and standard deviation.

So if any of the two increases then the MOE also increases, thus widening the confidence interval.

And the MOE is inversely related to the sample size.

So if the sample increases the MOE decreases and vice versa.

It is provided that the sample size and the sample proportion are not altered.

The critical value of z for 90% confidence level is:

$z_{\alpha/2}= z_{0.10/2}=z_{0.05}=1.645$

And the critical value of z for 99% confidence level is:

$z_{\alpha/2}= z_{0.05/2}=z_{0.05}=1.96$

So as the confidence level increases the critical value increases.

Thus, a 90% confidence interval for p will be narrower than the 99% confidence interval.

6 0

3 years ago

An e-mail filter is planned to separate valid e-mails from spam. The word free occurs in 60% of the spam messages and only 4% of

ANEK [815]

Answer:

(a) 0.152

(b) 0.789

Step-by-step explanation:

Let's denote the events as follows:

F = The word free occurs in an email

S = The email is spam

V = The email is valid.

The information provided to us are:

The probability of the word free occurring in a spam message is, $P(F|S)=0.60$
The probability of the word free occurring in a valid message is, $P(F|V)=0.04$
The probability of spam messages is,

$P(S)=0.20$

First let's compute the probability of valid messages:

$P (V) = 1 - P(S)\\=1-0.20\\=0.80$

(a)

To compute the probability of messages that contains the word free use the rule of total probability.

The rule of total probability is:

$P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})$

The probability that a message contains the word free is:

$P(F)=P(F|S)P(S)+P(F|V)P(V)\\=(0.60*0.20)+(0.04*0.80)\\=0.152\\$

The probability of a message containing the word free is 0.152

(b)

To compute the probability of messages that are spam given that they contain the word free use the Bayes' Theorem.

The Bayes' theorem is used to determine the probability of an event based on the fact that another event has already occurred. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is spam provided that it contains free is:

$P(S|F)=\frac{P(F|S)P(S)}{P(F)}\\=\frac{0.60*0.20}{0.152} \\=0.78947\\$

The probability that a message is spam provided that it contains free is approximately 0.789.

(c)

To compute the probability of messages that are valid given that they do not contain the word free use the Bayes' Theorem. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is valid provided that it does not contain free is:

$P(V|F^{c})=\frac{P(F^{c}|V)P(V)}{P(F^{c})} \\=\frac{(1-P(F|V))P(V)}{1-P(F)}\\=\frac{(1-0.04)*0.80}{1-0.152} \\=0.90566$

The probability that a message is valid provided that it does not contain free is approximately 0.906.

4 0

3 years ago

Congruent Triangles and Proofs Given: JK bisects

Nikitich [7]

Answer:

Step-by-step explanation:

1S. JM bisects ∠ KJL and ∠JMK ≅ ∠JML

1R. Given

2S. ∠KJM ≅ ∠LJM

2R. Angle bisectors form ≅ ∠s

3S. JM≅JM

3R. Reflexive Propriety ( segment is ≅ with itself)

4S. ΔJMK ≅ΔJML

4R. ASA theorem of congruency

5S. JK≅JL

5R. Coresponding parts of congruent Δs are congruent (CPCTC)

7 0

3 years ago