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NISA [10]
3 years ago
11

Solve for A a=22/7•13/4^2

Mathematics
2 answers:
Mkey [24]3 years ago
7 0

Answer:

A= ~2.55

Step-by-step explanation:

Using Pemdos, or gemdos, you can find you do the exponent first

4^2 = 16

then to everything else in left to right order since they're all multiplication or division

22/7 = ~3.14 * 13 = ~40.86/16 = ~2.55

(I'm using "~" to mean about/rounded to)

erik [133]3 years ago
7 0
A=2.55

Solve 4^2=16

If u wanna uses the calculator write it like this:
(22/7)•(13/16)

Then u get the answer
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Kevin has 3 fewer leaves than sandy. Sandy has 3 leaves. How many leaves does Kevin have?
chubhunter [2.5K]

Answer:

0 leaves

Step-by-step explanation:

6 0
3 years ago
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If the distance covered by an object in time t is given by s(t)=t^2+5t , where s(t) is in meters and t is in seconds, what is th
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Answer:

E. 44 meters

Step-by-step explanation:

The function that models the distance covered by the object is

s(t)=t^2+5t

where s(t) is in meters and t is in seconds.

The distance covered by the object after 1 second is

s(1)=1^2+5(1)=6m

The distance covered by the object after 1 second is

s(1)=5^2+5(5)=50m

The distance covered between 1 second and 5 seconds is

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6 0
3 years ago
Ten years ago 53% of American families owned stocks or stock funds. Sample data collected by the Investment Company Institute in
Alborosie

Answer:

a) Null hypothesis:p\geq 0.53  

Alternative hypothesis:p < 0.53  

b) z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429  

p_v =P(Z

c) So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .  

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

\hat p=0.46 estimated proportion of American families owning stocks or stock funds

p_o=0.53 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

Part a

We need to conduct a hypothesis in order to test the claim that proportion is less than 0.53 or 53%.:  

Null hypothesis:p\geq 0.53  

Alternative hypothesis:p < 0.53  

Part b

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z

Part c  

So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .  

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To answer the question above, we are simply to subtract the length of the gold ribbon which is 2 4/6 ft from the length of the silver ribbon, 5 2/6 feet. Mathematically,
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3 years ago
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