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3 years ago

Which statement best describes the equation x5 + x3 – 14 = 0?

Mathematics

2 answers:

damaskus [11]3 years ago

4 0

Answer:

2.The equation is not quadratic in form because it cannot be rewritten as a second-degree polynomial.

Step-by-step explanation:

A quadratic equation is any equation that can be rearranged in standard form as :

ax² + bx + c = 0

Where a, b and c are coefficients and a ≠ 0.

Since for a quadratic equation, the power of x is a non negative integer, it is considered as a polynomial. A quadratic equation is a second-degree polynomial (i.e the gratest power of x is two).

The equation is not a quadratic equation because it cannot be rewritten as a second-degree polynomial.

MrRissso [65]3 years ago

4 0

Answer:

B). The equation is quadratic in form because the difference of the exponent of the lead term and the exponent of the middle term is 2.

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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

1.5d+3=-4.5 what does D equal

vagabundo [1.1K]

Answer:

d = − 5

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

7 0

3 years ago

Read 2 more answers

Factor the expression using the GCF 42-12

Monica [59]

6(7 - 2)
6(5)

Check:
7*6 = 42
2*6 = 12
42-12 = 30 = 6(5) :)

5 0

3 years ago

A line has a slope of zero and passes through the point (-2, 3). Which of the following points must also lie on the line?

brilliants [131]

Answer:

(5,3)

Step-by-step explanation

Because the slope is 0, it means its a horizontal line. because the horizontal is on 3, then it doesnt matter what x is as long as y is 3 if that makes any sense.

7 0

3 years ago

Use spherical coordinates. evaluate (9 − x2 − y2) dv, where h is the solid hemisphere x2 + y2 + z2 ≤ 4, z ≥ 0.

avanturin [10]

In spherical coordinates, we have

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}$

which gives volume element

$\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

and so the triple integral is given by

$\displaystyle\iiint_H(9-x^2-y^2)\,\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/2}\int_{\rho=0}^{\rho=2}(9-\rho^2\sin^2\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$
$=\displaystyle2\pi\int_{\varphi=0}^{\varphi=\pi/2}\int_{\rho=0}^{\rho=2}(9\rho^2\sin\varphi-\rho^4\sin^3\varphi)\,\mathrm d\rho\,\mathrm d\varphi$
$=\dfrac{592\pi}{15}$

4 0

3 years ago