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Afina-wow [57]
3 years ago
15

What is 0.86/5-0.3*0.5

Mathematics
2 answers:
klio [65]3 years ago
7 0
0.86/5=0.172
0.3×0.5=0.15
0.172_0.15=0.022
grigory [225]3 years ago
5 0
0.86/5-0.3*0.5= 2 answers

1. can be 11/500

2. it can be 0.022 but im positive it has to be 11/500
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Pluto is 5,560,000,000 years old. Estimate the age of Pluto by writing it in the form of x×10y years, where x and y are single-d
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Answer:

The age of pluto is:

5,560,000,000 years old = 5.56 × 10⁹ years old

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Step-by-step explanation:

Pluto is 5,560,000,000 years old. Estimate the age of Pluto by writing it in the form of x×10y years, where x and y are single-digit numbers.

We write this in standard form or scientific notation.

5,560,000,000 years old = 5.56 × 10⁹ years old

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4 0
3 years ago
A theory predicts that the mean age of stars within a particular type of star cluster is 3.3 billion years, with a standard devi
Luden [163]

Answer:

Null hypothesis:\mu \leq 3.3  

Alternative hypothesis:\mu > 3.3  

z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768  

Since is a one right tailed test the p value would be:  

p_v =P(z>1.768)=0.039  

Step-by-step explanation:

Data given and notation

\bar X=3.4 represent the sample mean  

\sigma=0.4 represent the population deviation for the sample

n=50 sample size  

\mu_o =3.3 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean is higher than 3.3, the system of hypothesis would be:  

Null hypothesis:\mu \leq 3.3  

Alternative hypothesis:\mu > 3.3  

Compute the test statistic  

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768  

P value

Since is a one right tailed test the p value would be:  

p_v =P(z>1.768)=0.039  

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