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amid [387]
3 years ago
13

How many roots does the graph polynomial function have?

Mathematics
2 answers:
lisov135 [29]3 years ago
6 0
The graph crosses the x axis 3 times so there are 3 roots.

B
sergejj [24]3 years ago
3 0

Answer:

The number of roots of the graph is 3

B is correct.

Step-by-step explanation:

We are given a graph of polynomial.

Graph start from negative infinity and end at positive infinity.

The degree of polynomial must be odd.

In graph, we can see it cuts at three points

(-6,0)    (-2,0)    (-1,0)

This graph has three x-intercept.

Number of x-intercept is equivalent to real roots of the polynomial.

This graph is cubic polynomial because it has three x-intercept.

Hence, The number of roots of the graph is 3

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The cone and the cylinder below have equal surface area. True or false.
WARRIOR [948]

Answer:

False

Step-by-step explanation:

The surface area of the cone is

V=\pi r^2 +\pi rl

SA=\pi r^2 +\pi\times r\times 2r

SA=\pi r^2 +2\pi\times r^2

SA=3\pi r^2

The surface area of the cylinder is:

SA=2\pi r^2 +2\pi rh

SA=2\pi r^2 +2\pi\times r\times r

SA=4\pi r^2

3 0
3 years ago
(b) The area of a rectangular painting is 7719 cm2.
Nataliya [291]

Given:

  • The area of a rectangular painting is 7719 cm²
  • Width of painting = 83cm

To Find?

  • Length of painting.

Solution:

  • Let length of painting be x cm

Using formula:

  • Area of rectangle = L × B

Where,

  • Length = x cm
  • Breadth = 73 cm

→ 7719 = x × 83

→ x = 7719/83

→ x = 93 cm

Hence,

  • Length of painting is 93cm
6 0
2 years ago
Read 2 more answers
Simplify.<br> 2u? - 20u +42<br> u² – 5u+6
chubhunter [2.5K]

Answer:

The simplified expression is:

\frac{2(u + 7)}{(u- 2)}

Step-by-step explanation:

We want to simplify:

\frac{2 {u}^{2}  - 20u + 42}{ {u}^{2}  - 5u + 6}

We factor the numerator and the denominator to get:

\frac{2(u - 3)(u + 7)}{(u- 2)(u- 3)}

Cancel out u-2 which is common to the numerator and the denominator yo get:

\frac{2(u + 7)}{(u- 2)}

5 0
3 years ago
What is 0.18 repeating simplified
Radda [10]
As a fraction it’s, 2/11
8 0
3 years ago
Researchers studying the acquisition of pronunciation often compare measurements made on the recorded speech of adults and child
dolphi86 [110]

Answer:

(A) The additional information that is needed to confirm about the conditions for this test have been met is ‘Population is approximately normal.

(B) The test statistic = 1.9965

(C1) P-value = 0.0556

(C2) There is no significant difference between mean VOT for children and adults.

D1) It is possible to make type II error.

(D2) There should be a difference in the VOT of adults and children.

Step-by-step explanation:

Let na be the number of adults = 20

xa mean VOT for adults = 88.17

sa standard deviatiation of VOT for adults = 24.74

Let nc be the number of children = 10

xc mean VOT for children =60.67

sc standard deviatiation of VOT for children = 39.89

(A) From the information the population variances are unknown and the two sample are assumed to be independent and the sample the sample size are smaller that is (n<30).

This indicates that the additional information that is required for the conditions of the test to be satisfied is 'distribution of the population'. the addition assumption to be made is, that the population distribution is normal.

(b) Calculating the test statistics using the formula;

t = (xa -xc)/SE - d

where SE = standard deviation , d= hypothesized difference = 0

But SE = √sa²/na +sc²/nc

           = √24.74 ²/20 + 39.89²/10

           = √189.72559

           = 13.774

Substituting into test statistics equation, we have

t = (xa -xc)/SE - d

  = (88.17 - 60.67/13.774

  = 1.9965

Therefore the test statistic is 1.9965

(c) Calculating the p-value, we have;

Degree of freedom = na + nc -2

                                 = 10+ 20 -2

                                 = 28

The p-value for t=1.9965 at 28 degrees of freedom and 0.05 level of significance is 0.0556.

The p-value 0.0556 is greater than given level of significance 0.05 hence we fail to reject the null hypothesis and conclude that there is no significant difference between mean VOT for children and adults.

(D1) From the information in part (C) the null hypothesis is not rejected.

Since the null hypothesis is not rejected, there might be a chance that not rejecting null hypothesis would be wrong. In this type of situation the error that can occur would be type II error.

(D2) The type of error describe in the context of this study is obtained by the concept of the type II error which tells that the null hypothesis is not rejected when it is actually false.

7 0
3 years ago
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