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Helga [31]
3 years ago
7

Convert to gallon of 144 cup

Mathematics
1 answer:
Ivanshal [37]3 years ago
3 0
This is the answer to your question

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Step-by-step explanation:

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The partial pressure of oxygen PaO2 is a measure ofthe amount of oxygen in the blood. Assume that the distribution ofPaO2 levels
Rus_ich [418]

Answer: a) 0.8665

b) 0.8190

Step-by-step explanation:

Given : The partial pressure of oxygen PaO2 is a measure ofthe amount of oxygen in the blood. Assume that the distribution ofPaO2 levels among newborns has a \mu=38 mmHg and \sigma= 9 mmHg.

If we take a random sample n= 25 newborns, then  using formula z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}, we have

At x= 36

z=\dfrac{36-38}{\dfrac{9}{\sqrt{25}}}\approx-1.11

At x= 41

z=\dfrac{36-38}{\dfrac{9}{\sqrt{25}}}\approx1.67

Using table for z-values, the probability that the sample mean will be greater than 36 :

P(z>-1.11)=1-P(\leq-1.11)=1-(1-P(z\leq1.11))=P(z\leq1.11)=0.8665004\approx0.8665

The probability that the sample mean will be between 36 and 41 :-

P(-1.11

4 0
3 years ago
Which point on the graph of g(x)=(1/5)^x? HELPP
Cloud [144]

Answer:

(-1,5) and (3, \frac{1}{125}) are points on the graph

Step-by-step explanation:

Given

g(x) = \frac{1}{5}^x

Required

Determine which point in on the graph

To get which of point A to D is on the graph, we have to plug in their values in the given expression using the format; (x,g(x))

A. (-1,5)

x = -1

Substitute -1 for x in g(x) = \frac{1}{5}^x

g(x) = \frac{1}{5}^{-1}

Convert to index form

g(x) = 1/(\frac{1}{5})

Change / to *

g(x) = 1*(\frac{5}{1})

g(x) = 5

This satisfies (-1,5)

<em>Hence, (-1,5) is on the graph</em>

<em></em>

B. (1,0)

x = 1

Substitute 1 for x

g(x) = \frac{1}{5}^x

g(x) = \frac{1}{5}^1

g(x) = \frac{1}{5}

<em>(1,0) is not on the graph because g(x) is not equal to 0</em>

C. (3, \frac{1}{125})

x = 3

Substitute 3 for x

g(x) = \frac{1}{5}^x

g(x) = \frac{1}{5}^3

Apply law of indices

g(x) = \frac{1}{5} * \frac{1}{5} * \frac{1}{5}

g(x) = \frac{1}{125}

This satisfies (3, \frac{1}{125})

<em>Hence, </em>(3, \frac{1}{125})<em> is on the graph</em>

<em></em>

D.  (-2, \frac{1}{25})

x = -2

Substitute -2 for x

g(x) = \frac{1}{5}^x

g(x) = \frac{1}{5}^{-2}

Convert to index form

g(x) = 1/(\frac{1}{5}^2)

g(x) = 1/(\frac{1}{5}*\frac{1}{5})

g(x) = 1/(\frac{1}{25})

Change / to *

g(x) = 1*(\frac{25}{1})

g(x) = 25

This does not satisfy  (-2, \frac{1}{25})

<em>Hence, </em> (-2, \frac{1}{25})<em> is not on the graph</em>

8 0
4 years ago
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