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Alexeev081 [22]
3 years ago
11

X2 +4+5y3+2x2-2x2-6+1​

Mathematics
2 answers:
Andreyy893 years ago
6 0

Answer:

x^{3} +5y^{3} -1

Step-by-step explanation:

Verdich [7]3 years ago
6 0

Answer: x^2-1+5y^3

Step-by-step explanation:

x^2+4+5y^3-6+1 (2x^2-2x^2 cancel each other.

x^2-1+5y^3 (4-6+1= -1)

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Plz help I’m getting timed
bixtya [17]

Answer:

17in.

Step-by-step explanation:

7 0
2 years ago
An online streaming service providing television programs claims that a 30-minute program will stream with advertisements that a
trapecia [35]

Answer:

The p-value is significantly low(<0.01), which means that the data provides convincing evidence that the true mean advertisement length is longer than 45 seconds.

Step-by-step explanation:

An online streaming service providing television programs claims that a 30-minute program will stream with advertisements that average 45 seconds. Test if the true mean advertisement length is longer than 45 seconds.

At the null hypothesis, we test if the mean time is of 45 seconds, that is:

H_0: \mu = 45

At the alternate hypothesis, we test if the mean is more than 45 seconds, that is:

H_1: \mu > 45

The test statistic is:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

45 is tested at the null hypothesis:

This means that \mu = 45

They recorded the times of 21 randomly selected advertisements. The mean and standard deviation for these times are 46.67 and 2.78.

This means that n = 21, X = 46.67, s = 2.78

Value of the test-statistic:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{46.67 - 45}{\frac{2.78}{\sqrt{21}}}

t = 2.75

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 46.67, which is a right-tailed test, with t = 2.75 and 21 - 1 = 20 degrees of freedom.

With the help of a calculator, this p-value is of 0.0062.

The p-value is significantly low(<0.01), which means that the data provides convincing evidence that the true mean advertisement length is longer than 45 seconds.

6 0
3 years ago
Gina’s kitten weighs 5 ounces. Of the kitten gains 3 ounces every week, how many ounces will the kitten weigh in 6 weeks?
Alenkasestr [34]

Answer:

23 ounces

Step-by-step explanation:

3x6=18 +5 = 23

8 0
3 years ago
Read 2 more answers
In 2010, the average duration of long-distance telephone calls originating in one town was 9.4 minutes. A long-distance telephon
Zepler [3.9K]

<u>Answer</u>: No, we do not have sufficient evidence to conclude that the mean call duration, µ, is different from the 2010 mean of 9.4 minutes.

Step-by-step explanation:

As per given , we have

H_0: \mu=9.4\\\\H_a:\mu\neq9.4, since H_0 is two-tailed so , the test is a two tail test.

Since population standard deviation is unknown, so we use t-test.

Critical value (two-tailed) for significance level of 0.01= t_{n-1,\alpha/2}=t_{49, 0.005}\pm2.609228

For n =50 , \overline{x}=8.6 and s= 4.8

Test statistic : t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}

t=\dfrac{8.6-9.4}{\dfrac{4.8}{\sqrt{50}}}\approx-1.18

Since test statistic value (-1.18) lies in critical interval (-2.609228, 2.609228), it means the null hypothesis is failed to reject.

We do not have sufficient evidence to conclude that the mean call duration, µ, is different from the 2010 mean of 9.4 minutes.

4 0
2 years ago
Does anyone wanna be friends lol?
USPshnik [31]

Answer:

hahaha me

Step-by-step explanation:

...actually how old are you though lol

7 0
3 years ago
Read 2 more answers
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