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Levart [38]
3 years ago
11

Please Help Me!!

Mathematics
1 answer:
Lemur [1.5K]3 years ago
3 0
Perimeter: x + (4x + 9) + (x + 11)
The value x is for the bottom of the triangle.
Simplified, it would be 6x + 20
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Your load is a container full of water. The container measures
solmaris [256]

Answer:

1282kg

Step-by-step explanation:

1.0*1.12*1.1=1.232 tonnes

50kg + 1232kg = 1282kg

Hope this helps :)

7 0
2 years ago
Read 2 more answers
Find f'(x) for f(x)=cos^2(5x^3).
-BARSIC- [3]

d/dx cos^2(5x^3)

= d/dx [cos(5x^3)]^2

= 2[cos(5x^3)]

= - 2[cos(5x^3)] * sin(5x^3)

= - 2[cos(5x^3)] * sin(5x^3) * 15x^2

= - 30[cos(5x^3)] * sin(5x^3) * x^2

Explanation:

d/dx x^n = nx^(n - 1)

d/dx cos x = - sin x

Chain rule:

d/dx f(g(...w(x))) = f’(g(...w(x))) * g’(...w(x)) * ... * w’(x)

6 0
3 years ago
Read 2 more answers
Please help is it 2/9?​
attashe74 [19]

Answer:

7/9    

Step-by-step explanation:

7/9

4 0
2 years ago
Read 2 more answers
A recent study reported that high school students spend an average of 94 minutes per day texting. Jenna claims that the average
larisa [96]

Answer:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

Step-by-step explanation:

Jenna claims that the average time of texting at her larger high school is greater than 94 minutes per day.

From here we can see that we have to perform a hypothesis test about a sample mean. The null and alternate hypothesis will be:

Null Hypothesis: \mu \leq  94

Alternate Hypothesis: \mu > 94

Jenna collected data from a sample of 32 students. So, sample size will be:

Sample Size = n = 32

Sample Mean = x = 96.5

Sample Standard Deviation = s = 6.3

We have to perform a hypothesis test, to test Jenna's claim. Since, the value of Population Standard Deviation is unknown and the value of Sample Standard Deviation is known, we will use One Sample t-test in this case.

The formula to calculate the test statistic is:

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

Using the values, we get:

t=\frac{96.5-94}{\frac{6.3}{\sqrt{32} } }=2.245

The degrees of freedom will be:

df = n - 1 = 32 - 1 = 31

We have to convert the t-score 2.245 with 31 degrees of freedom to its equivalent p-value. From t-table this value comes out to be:

p-value = 0.0160

The significance level is:

\alpha =0.05

Since, the p-value is lesser than the level of significance, we reject the Null Hypothesis.

Conclusion:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

3 0
3 years ago
Malcolm is downloading a file that is 50.4 Mb. his internet connection has a download speed of 1.2 mb per second. Let t represen
VikaD [51]
The equation is 1.2t=50.2 And the solution is 42
5 0
3 years ago
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