What set of reflections would carry hexagon ABCDEF onto itself?

Un padre para estimular a su hijo para estudiar matemáticas le dice: por cada ejercicio que resuelvas bien te daré 7 € y por cad

Naya [18.7K]

Answer:

8 ejercicios

Step-by-step explanation:

De acuerdo a la información dada, puedes plantear las siguientes ecuaciones:

x+y= 45 (1)

7x-1y= 19 (2)

x= ejercicios que resolvió bien

y= ejercicios que resolvió mal

Primero, puedes despejar x en (1):

x= 45-y (3)

Segundo, debes reemplazar (3) en (2):

7(45-y)-1y=19

315-7y-1y= 19

315-8y= 19

315-19= 8y

296= 8y

y= 296/8

y= 37

Tercero, debes reemplazar el valor de y en (3) para encontrar x:

x= 45-37

x= 8

De acuerdo a esto, la respuesta es que el hijo ha resuelto bien 8 ejercicios.

8 0

3 years ago

Several students tested plant growth for given conditions. Which student’s data is most representative of average plant growth f

Roman55 [17]

Brenden. the more plants you test the easier it is to find the average instead of a smaller amount. hope this helped!

3 0

3 years ago

Read 2 more answers

62% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 48 owned dogs are r

dedylja [7]

Answer:

a) 0.1180 = 11.80% probability that exactly 30 of them are spayed or neutered.

b) 0.8665 = 86.65% probability that at most 33 of them are spayed or neutered.

c) 0.4129 = 41.29% probability that at least 31 of them are spayed or neutered.

d) 0.5557 = 55.57% probability that between 24 and 30 of them are spayed or neutered.

Step-by-step explanation:

To solve this question, we use the binomial probability distribution, and also it's approximation to the normal distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

62% of owned dogs in the United States are spayed or neutered.

This means that $p = 0.62$

48 owned dogs are randomly selected

This means that $n = 48$

Mean and standard deviation, for the approximation:

$\mu = E(x) = np = 48*0.62 = 29.76$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{48*0.62*0.38} = 3.36$

a. Exactly 30 of them are spayed or neutered.

This is P(X = 30), which is not necessary the use of the approximation.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 30) = C_{48,30}.(0.62)^{30}.(0.38)^{18} = 0.1180$

0.1180 = 11.80% probability that exactly 30 of them are spayed or neutered.

b. At most 33 of them are spayed or neutered.

Now we use the approximation. This is, using continuity correction, $P(X \leq 33 + 0.5) = P(X \leq 33.5)$ , which is the pvalue of Z when X = 33.5. So

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

$Z = \frac{33.5 - 29.76}{3.36}$

$Z = 1.11$

$Z = 1.11$ has a pvalue of 0.8665

0.8665 = 86.65% probability that at most 33 of them are spayed or neutered.

c. At least 31 of them are spayed or neutered.

Using continuity correction, this is $P(X \geq 31 - 0.5) = P(X \geq 30.5)$ , which is 1 subtracted by the pvalue of Z when X = 30.5. So

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

$Z = \frac{30.5 - 29.76}{3.36}$

$Z = 0.22$

$Z = 0.22$ has a pvalue of 0.5871

1 - 0.5871 = 0.4129

0.4129 = 41.29% probability that at least 31 of them are spayed or neutered.

d. Between 24 and 30 (including 24 and 30) of them are spayed or neutered.

This is, using continuity correction, $P(24 - 0.5 \leq X \leq 30 + 0.5) = P(23.5 \leq X \leq 30.5)$ , which is the pvalue of Z when X = 30.5 subtracted by the pvalue of Z when X = 23.5.

X = 30.5

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

$Z = \frac{30.5 - 29.76}{3.36}$

$Z = 0.22$

$Z = 0.22$ has a pvalue of 0.5871

X = 23.5

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

$Z = \frac{23.5 - 29.76}{3.36}$

$Z = -1.86$

$Z = -1.86$ has a pvalue of 0.0314

0.5871 - 0.0314 = 0.5557

0.5557 = 55.57% probability that between 24 and 30 of them are spayed or neutered.

8 0

3 years ago

The director of admissions at Kinzua University in Nova Scotia estimated the distribution of student admissions for the fall sem

Sophie [7]

Answer:

Variance =10900.00

Standard deviation=104.50

Step by step Explanation:

Admissions Probability for 1100= 0.2

Admissions Probability for 1400=0.3

Admissions Probability for 1300 =0.5

To find the expected value, we will multiply each possibility by its probability and then add.

mean = 1100*0.2 + 1400*0.3 + 1300*0.5 = 1290

To find the variance, we will start by squaring each possibility and then multiplying it by its probability. We will then add these and subtract the mean squared.

E(X^2)=( 1100²*0.2)+ (1400²*0.3 )+ (1300²*0.5) = 1675000

Variance(X)=E(X²)- [E(X)]²

= 1675000 - (1290)²

=10900

Hence, the Variance(X)=10900

Then to calculate the standard variation , we will use the formular below,

standard variation (X)=√ var(X)= √10900

=104.5

Hence the standard variation=104.5

8 0

3 years ago

Which expression is equivalent to - 14 - 6?

Charra [1.4K]

The answer is d. -14 + (-6)

8 0

3 years ago