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goblinko [34]
3 years ago
14

The population of flamingos at the Bronx Zoo is 48.One spring, they have lots of babies and the population increases by 12.5%.Ho

w many flamingos are there now .​
Mathematics
1 answer:
mihalych1998 [28]3 years ago
4 0
The answer you are looking for is 54.
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20 POINTS Compost costs $10 per cubic yard and topsoil costs $2 per cubic yard. Maria needs more than 10 cubic feet of topsoil.
Morgarella [4.7K]
I can't see the graph but let's use logic
hmm, more than 10 cubic feet of topsoil so the first and 2nd options are wrong

let's ee the costs
3rd option
10*1=10
2*12=24
10+24=34 and 34<50, that is fine

4th option
3*10=30
2*12=24
30+24=54
54>50, nope, that is over cost


answer is 3rd one
the one with 1 cubic yard compost and 12 cubic yard topsoil
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Answer:

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Step-by-step explanation:

Disagree, it is not always correct

Consider dividing 5 by 0.5, you get 10 which is greater than what you started with

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Step-by-step explanation:

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What is X to the third power plus 125?
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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 .  

7 0
3 years ago
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