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3 years ago

Plz help will mark brainliest

Mathematics

2 answers:

vladimir1956 [14]3 years ago

7 0

Answer:

It is fuzzy but it looks like ∠A is 35°. If that's correct, then:

∠D = 35°

(if ∠A is 36°, then ∠D is 36°)

Step-by-step explanation:

It is indicated that ∠C and ∠F are 90°.

It is also indicated that ∠B and ∠E are equal.

Therefore, ∠A = ∠D

Mama L [17]3 years ago

3 0

Answer:

the correct answer is <D is 72 degrees

Step-by-step explanation:

x + 36=180°

x= 180-36

x=144°

To find angle d divide your answer by 2

therefore angle d is 72°

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If m (x) = StartFraction x + 5 Over x minus 1 EndFraction and n(x) = x – 3, which function has the same domain as (m circle n) (

Mars2501 [29]

Question:

If m (x) = StartFraction x + 5 Over x minus 1 EndFraction and n(x) = x – 3, which function has the same domain as (m circle n) (x)?

h (x) = StartFraction x + 5 Over 11 EndFraction

h (x) = StartFraction 11 Over x minus 1 EndFraction

h (x) = StartFraction 11 Over x minus 4 EndFraction

h (x) = StartFraction 11 Over x minus 3 EndFraction

Answer:

Option C: $h(x)=\frac{11}{x-4}$ has the same domain as $(m \circ n)(x)$

Explanation:

It is given that $m(x)=\frac{x+5}{x-1}$ and $n(x)=x-3$

Let us find the domain of $(m \circ n)(x)$

$\begin{aligned}(m \circ n)(x) &=m(n(x))\\&=m(x-3) \\ &=\frac{(x-3)+5}{(x-3)-1} \\ &=\frac{x+2}{x-4} \end{aligned}$

Now, let us equate the denominator equal to zero to determine the domain.

$x-4=0$

$x=4$

Thus, the function becomes undefined at the point $x=4$

Hence, the domain of $(m \circ n)(x)$ is $(-\infty, 4) \cup(4, \infty)$

Now, we shall find the function which has the same domain as $(m \circ n)(x)$

Option A: $h(x)=\frac{x+5}{11}$

The function h(x) has the domain of set of all real numbers $$-\infty$

Thus, the interval $(-\infty,\infty)$ is not the same domain as $(-\infty, 4) \cup(4, \infty)$

Hence, Option A is not the correct answer.

Option B: $h(x)=\frac{11}{x-1}$

Equating the denominator equal to zero, the function becomes undefined at the point $x=1$

Thus, the function h(x) has the domain of $(-\infty,-1) \cup(-1, \infty)$ is not the same domain as $(-\infty, 4) \cup(4, \infty)$

Hence, Option B is not the correct answer.

Option C: $h(x)=\frac{11}{x-4}$

Equating the denominator equal to zero, the function becomes undefined at the point $x=4$

Thus, the function h(x) has the domain of $(-\infty, 4) \cup(4, \infty)$ is the same domain as $(-\infty, 4) \cup(4, \infty)$

Hence, Option C is the correct answer.

Option D: $h(x)=\frac{11}{x-3}$

Equating the denominator equal to zero, the function becomes undefined at the point $x=3$

Thus, the function h(x) has the domain of $(-\infty,3) \cup(3, \infty)$ is not the same domain as $(-\infty, 4) \cup(4, \infty)$

Hence, Option D is not the correct answer.

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3 years ago

Simplity -12 +33-(3+4) + (5)

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Answer:

Step-by-step explanation:

−12+33−(3+4)+5

=21−(3+4)+5

=21−7+5

=14+5

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3 years ago

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Simply 3x +9y + 2x<br> Please help me

dem82 [27]

Answer:

5x+9y

Step-by-step explanation:

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3 years ago

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The amount of warpage in a type of wafer used in the manufacture of integrated circuits has mean 1.3 mm and standard deviation 0

valina [46]

Answer:

a) $P(\bar X >1.305)=P(Z>\frac{1.305-1.3}{\frac{0.1}{\sqrt{200}}}=0.707)$

And using the complement rule, a calculator, excel or the normal standard table we have that:

$P(Z>0.707)=1-P(Z$

b) $z=-0.674$

And if we solve for a we got

$a=1.3 -0.674* \frac{0.1}{\sqrt{200}}=$

So the value of height that separates the bottom 95% of data from the top 5% is 1.295.

c) $P( \bar X >1.305) = 0.05$

We can use the z score formula:

$P( \bar X >1.305) = 1-P(\bar X$

Then we have this:

$P(z< \frac{1.305-1.3}{\frac{0.1}{\sqrt{n}}}) = 0.95$

And a value that accumulates 0.95 of the area on the normal distribution z = 1.64 and we can solve for n like this:

$n = (1.64*\frac{0.1}{1.305-1.3})^2= 1075.84 \approx 1076$

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

Let X the random variable that represent the amount of warpage of a population and we know

Where $\mu=1.3$ and $\sigma=0.1$

Since the sample size is large enough we can use the central limit theorem andwe know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

We can find the probability required like this:

$P(\bar X >1.305)=P(Z>\frac{1.305-1.3}{\frac{0.1}{\sqrt{200}}}=0.707)$

And using the complement rule, a calculator, excel or the normal standard table we have that:

$P(Z>0.707)=1-P(Z$

Part b

For this part we want to find a value a, such that we satisfy this condition:

$P(\bar X>a)=0.75$ (a)

$P(\bar X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.674$

And if we solve for a we got

$a=1.3 -0.674* \frac{0.1}{\sqrt{200}}=$

So the value of height that separates the bottom 95% of data from the top 5% is 1.295.

Part c

For this case we want this condition:

$P( \bar X >1.305) = 0.05$

We can use the z score formula:

$P( \bar X >1.305) = 1-P(\bar X$

Then we have this:

$P(z< \frac{1.305-1.3}{\frac{0.1}{\sqrt{n}}}) = 0.95$

And a value that accumulates 0.95 of the area on the normal distribution z = 1.64 and we can solve for n like this:

$n = (1.64*\frac{0.1}{1.305-1.3})^2= 1075.84 \approx 1076$

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3 years ago

Bonds from the U.S. Treasury are selling at 102.293, bonds from Alachua County are selling at 85.681, and bonds from the city of

Anestetic [448]

Answer:

a. $3000

Step-by-step explanation:

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