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qaws [65]
3 years ago
11

Mike bought a soft drink for 4 dollars and 9 candy bars. He spent a total of 31 dollars. How much did each candy bar cost ?

Mathematics
1 answer:
Andre45 [30]3 years ago
7 0

Answer:

$3 per candy bar

Step-by-step explanation:

:))

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The box part of the box plot contains all the values between which numbers?
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5 0
3 years ago
A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 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.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

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

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

5 0
3 years ago
Plz help me!!!
Nadya [2.5K]
To solve this we are going to use the exponential function: f(t)=a(1(+/-)b)^t
where
f(t) is the final amount after t years
a is the initial amount
b is the decay  or grow rate rate in decimal form
t is the time in years

Expression A 
f(t)=624(0.95)^{4t}
Since the base (0.95) is less than one, we have a decay rate here.
Now to find the rate b, we are going to use the formula: b=|1-base|*100%
b=|1-0.95|*100%
b=0.05*100%
b=5%
We can conclude that expression A decays at a rate of 5% every three months.

Now, to find the initial value of the function, we are going to evaluate the function at t=0
f(t)=624(0.95)^{4t}
f(0)=624(0.95)^{0t}
f(0)=624(0.95)^{0}
f(0)=624(1)
f(0)=624
We can conclude that the initial value of expression A is 624.

Expression B
f(t)=725(1.12)^{3t}
Since the base (1.12) is greater than 1, we have a growth rate here.
To find the rate, we are going to use the same equation as before:
b=|1-base|*100%
b=|1-1.12|*100
b=|-0.12|*100%
b=0.12*100%
b=12%
We can conclude that expression B grows at a rate of 12% every 4 months.

Just like before, to find the initial value of the expression, we are going to evaluate it at t=0
f(t)=725(1.12)^{3t}
f(0)=725(1.12)^{0t}
f(0)=725(1.12)^{0}
f(0)=725(1)
f(0)=725
The initial value of expression B is 725.

We can conclude that you should select the statements:
- Expression A decays at a rate of 5% every three months, while expression B grows at a rate of 12% every fourth months. 

- Expression A has an initial value of 624, while expression B has an initial value of 725.

8 0
3 years ago
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