All categories

3 years ago

PLEASE HELP QUICK!!!! What is the degree of 7x^6 - 6x^5 + 2x^3 + x - 8? A. 8 B. 7 C. 5 D. 6

1 answer:

5 0

For determining the correct degree of a polynomial, you need to look at each term and find the degree by adding the exponents of each variable in it. The largest degree is the degree of the polynomial.

7x^6 has largest degree of the given polynomial, so your answer is D.6

You might be interested in

HELP ASAP. Worth 10 points

finlep [7]

Answer:

They studied for a little more than an hour?

Step-by-step explanation:

7/12 of 60 Mins is 35 mins and 2/3 is 40 mins 40+35=75 mins

which is a little more than an hour, 15 mins more to be precise.

3 0

2 years ago

Solve the quadratic equation. Show all of your steps. x 2 + 3 x − 5 = 0

AlladinOne [14]

For this case we have the following quadratic equation:

$x ^ 2 + 3x-5 = 0$

Where:

$a = 1\\b = 3\\c = -5$

We look for the solution using the following formula:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

Substituting the values:

$x = \frac {-3 \pm \sqrt {3 ^ 2-4 (1) (- 5)}} {2 (1)}\\x = \frac {-3 \pm \sqrt {9 + 20}} {2}\\x = \frac {-3 \pm \sqrt {29}} {2}$

Thus, we have two roots:

$x_ {1} = \frac {-3+ \sqrt {29}} {2}\\x_ {2} = \frac {-3- \sqrt {29}} {2}$

Answer:

$x_ {1} = \frac {-3+ \sqrt {29}} {2}\\x_ {2} = \frac {-3- \sqrt {29}} {2}$

3 0

3 years ago

Don’t know how to do this!

hram777 [196]

U+(v-t)
that’s the answer!

8 0

3 years ago

Read 2 more answers

To test Upper H 0: muequals50 versus Upper H 1: muless than50, a random sample of size nequals23 is obtained from a populatio

natta225 [31]

Answer:

Step-by-step explanation:

Hello!

<em>To test H0: u= 50 versus H1= u < 50, a random sample size of n = 23 is obtained from a population that is known to be normally distributed. Complete parts A through D. </em>

<em> A) If ¯ x = 47.9 and s=11.9, compute the test statistic .</em>

For thistest the corresponsing statistis is a one sample t-test

$t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }~~t_{n-1}$

$t_{H_0}= \frac{47.9-50}{\frac{11.9}{\sqrt{23} } } = -0.846= -0.85$

B) If the researcher decides to test this hypothesis at the a=0.1 level of significance, determine the critical value(s).

This test is one-tailed to the left, meaning that you'll reject the null hypothesis to small values of the statistic. The ejection region is defined by one critical value:

$t_{n-1;\alpha }= t_{22;0.1}= -1.321$

Check the second attachment. The first row shows α= Level of significance; the First column shows ν= sample size.

The t-table shows the values of the statistic for the right tail. P(tₙ≥α)

But keep in mind that this distribution is centered in zero, meaning that the right and left tails are numerically equal, only the sign changes. Since in this example the rejection region is one-tailed to the left, the critical value is negative.

C) What does the distribution graph appear like?

Attachment.

D) Will the researcher reject the null hypothesis?

As said, the rejection region is one-tailed to the right, so the decision rule is:

If $t_{H_0}$ ≤ -1.321, reject the null hypothesis.

If $t_{H_0}$ > -1.321, do not reject the null hypothesis.

$t_{H_0}$ = -0.85, the decision is to not reject the null hypothesis.

To test H0: μ=100 versus H1:≠100, a simple random sample size of nequals=24 is obtained from a population that is known to be normally distributed. Answer parts (a)-(d).

a) If x =104.2 and s=9.6, compute the test statistic.

For this example you have to use a one sample t-test too. The formula of the statistic is the same:

$t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } = \frac{104.2-100}{\frac{9.6}{\sqrt{24} } = } = 2.143$

b) If the researcher decides to test this hypothesis at the α=0.01 level of significance, determine the critical values.

This hypothesis pair leads to a two-tailed rejection region, meaning, you'll reject the null hypothesis at either small or big values of the statistic. Then the rejection region is divided into two and determined by two critical values (the left one will be negative and the right one will be positive but the module of both values will be equal).

$t_{n-1;\alpha/2 }= t_{23; 0.005}= -2.807$

$t_{n-1;1-\alpha /2}= t_{23;0.995}= 2.807$

c) Draw a t-distribution that depicts the critical region(s). Which of the following graphs shows the critical region(s) in thet-distribution?

Attachment.

(d) Will the researcher reject the null hypothesis?

The decision rule for the two-tailed hypotheses pair is:

If $t_{H_0}$ ≤ -2.807 or if $t_{H_0}$ ≥ 2.807, reject the null hypothesis.

If -2.807 < $t_{H_0}$ < 2.807, do not reject the null hypothesis.

$t_{H_0}$ = 2.143 is greater than the right critical value, the decision is to reject the null hypothesis.

Correct option:

B. The researcher will reject the null hypothesis since the test statistic is not between the critical values.

Full text in attachment. The sample size is different by 2 but it should serve as a good example.

H₀: μ = 20

H₁: μ < 20

a) n= 18, X[bar]= 18.3, S= 4, Compute statistic.

$t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }= \frac{18.3-20}{\frac{4}{\sqrt{18} } } = -1.80$

b) The rejection region in this example is one-tailed to the left, meaning that you'll reject the null hypothesis to small values of t.

Out of the three graphics, the correct one is A.

To resolve this you have to look for the values in the t-table that are the closest to the calculated $t_{H_0}$

Symbolically:

$t_{n-1;\alpha_1 } \leq t_{H_0}\leq t_{n-1;\alpha _2}$

$t_{H_0}$ = -1.80

$t_{17; 0.025 }= -2.110$

$t_{17;0.05}= -1.740$

Roughly defined you can say that the p-value is the probability of obtaining the value of $t_{H_0}$ , symbolically: P(t₁₇≤-1.80)

Under the distribution the calculated statistic is between the values of -2.110 and -1.740, then the p-value will be between their cumulated probabilities:

A. 0.025 < p-value < 0.05

d. The researcher decides to test the hypothesis using a significance level of α: 0.05

Using the p-value approach the decision rule is the following:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

We already established in item c) that the p-value is less than 0.05, so the decision is to reject the null hypothesis.

Correct option:

B. The researcher will reject the null hypothesis since the p-value is less than α.

I hope this helps!

6 0

3 years ago

A parabola can be drawn given a focus of (5, -3) and a directrix of y=1. Write the equation of the parabola in any form.

Anna71 [15]

Answer:

(x²-10x+33)/(-8) = y

Step-by-step explanation:

The distance between any point on a parabola from both its focus and directrix are the same.

Let's say we have a point (x,y) on the parabola. We can then say that using the distance formula,

$\sqrt{(x-5)^2+(y-(-3))^2}$ is the distance between (x,y) and the focus. Similarly, the distance between (x,y) and the directrix is |y-1| (I use absolute value here because distance is always positive). We can find this equation by taking the shortest distance from the point to the line. Because the closest point to the line will be the same x value as the point itself, the distance is simply the distance between the y value of the point and the y value of the directrix.

Equating the two equations given, we have

$\sqrt{(x-5)^2+(y-(-3))^2} = |y-1|$

square both sides to get

(x-5)²+(y+3)²=(y-1)²

expand the y components

(x-5)² + y²+6y+9 = y²-2y+1

subtract y²+6y+9 from both sides

(x-5)² = -8y - 8

expand the x components

x²-10x+25 = -8y - 8

add 8 to both sides to isolate the -8y

x²-10x+33 = -8y

divide both sides by -8 to isolate y

(x²-10x+33)/(-8) = y

6 0

3 years ago