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Sonja [21]
3 years ago
9

Please helppppppppppppppp

Mathematics
1 answer:
xz_007 [3.2K]3 years ago
7 0
6-72
step-by-step-explanation
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Each circle counts as one whole. Write a mixed number giving the amount shaded. Then, write this amount as an improper fraction.
ryzh [129]

Answer: 1 3/4 is your mixed fraction and 7/4 is the improper fraction.

Step-by-step explanation:

6 0
2 years ago
The graph below shows the height an elevator travels, y, in x seconds:
ad-work [718]

9514 1404 393

Answer:

  25 (ft/s)

Step-by-step explanation:

The "rate of change" is taken to mean the "rise" divided by the "run". Here, adjacent points differ in their y-value by 25 ft and their x-value by 1 second.

The rate of change is ...

  rise/run = (25 ft)/(1 second) = 25 ft/second

Usually, we're just interested in the numerical value: 25.

_____

<em>Additional comment</em>

The graph is mis-labeled. It is not showing "rate of ascent." It is showing height as a function of time. The "rate of ascent" is the <em>slope</em> of the line on the graph, not its <em>value</em>.

6 0
2 years ago
Assume that y varies inversely with x. If y=9 when x=7, find y when x=2.
Aleks [24]

Answer:1.56

Step-by-step explanation:

For constant of proportionality of k,

y=(1/k)*x

If y =9,

9=7/k

K=7/9=0.78

y=0.78x

y=0.78*2=1.56

When x

4 0
3 years ago
What is the value of -44
garik1379 [7]
Hi friend!


I presume you mean "What is the absolute value of -44!

Well an absolute value is always positive. So the absolute value of -44 is 44!


Hope I helped!


If not, tell me more clearly what you need!
4 0
3 years ago
Read 2 more answers
The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co
Alona [7]

Answer:

95% Confidence interval for the variance:

3.6511\leq \sigma^2\leq 34.5972

95% Confidence interval for the standard deviation:

1.9108\leq \sigma \leq 5.8819

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

s^2=2.89^2=8.3521

The confidence interval for the variance is:

\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128

Then, the confidence interval can be calculated as:

\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819

6 0
3 years ago
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