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Pavlova-9 [17]
4 years ago
12

The "prime time" for auditory and visual development is around what age?

Mathematics
2 answers:
Varvara68 [4.7K]4 years ago
8 0
C. 4 to 5 years

Hope this helps :)
7nadin3 [17]4 years ago
4 0

Answer:

The "prime time" for auditory and visual development is around:

c. 4 to 5 years

Step-by-step explanation:

Auditory and visual development in children refers to the development of capacities that allow them to perceive their surrondings and answer to them. The prime time to develop the capacity to hear and see is around 4 to 5 years.

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Is -7/3 equivalent to -7/-3?
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no

Step-by-step explanation:

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A rancher is ordering a box of cube-shaped salt licks. the edge length of each salt lick are 5/12 foot. is the volume of one sal
strojnjashka [21]

Answer:

0.0723 cubic feet < 1 cubic foot.

Step-by-step explanation:

A cube-shaped salt lick is ordered by a rancher and the length of each salt lick is \frac{5}{12} foot.

Therefore, the side length of each salt lick of cube-shape is \frac{5}{12} foot.

So, the volume of each salt lick is (\frac{5}{12} \times \frac{5}{12} \times \frac{5}{12}) = \frac{125}{1728} = 0.0723 cubic feet.

Hence, this volume is less than 1 cubic foot. (Answer)

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3 years ago
Identify the variable expression that is not a polynomial.
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I think the answer is A.33 not a polunomial
6 0
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What is the measure of the unknown acute angle in this right triangle?
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Answer:

The answer is 54

Step-by-step explanation:

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7 0
3 years ago
A sample of 1200 computer chips revealed that 45% of the chips fail in the first 1000 hours of their use. The company's promotio
yaroslaw [1]

Answer:

z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08

p_v = P(Z

So the p value obtained was a low value and using the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.   

Step-by-step explanation:

Data given and notation

n=1200 represent the random sample taken

\hat p=0.45 estimated proportion of chips that fail in the first 1000 hours of their use

\mu_0 =0.48 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion si less then 0.48:  

Null hypothesis:p\geq 0.48  

Alternative hypothesis:p < 0.48  

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  is significantly different from a hypothesized value .

Calculate the statistic  

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

z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08

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=0.05. 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

So the p value obtained was a low value and using the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.  

6 0
4 years ago
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