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Studentka2010 [4]
3 years ago
12

Solve the equation. 3/2x -5 = -11

Mathematics
2 answers:
zlopas [31]3 years ago
6 0
Answer: -4

Reasoning: To get rid of a fraction you need to multiply it by its inverse

ValentinkaMS [17]3 years ago
4 0

I think it's -4

If it's wrong sorry

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a beekeeper estimates that his bee population will triple each year. Currently he has 150 bees. Write a function to represent th
Soloha48 [4]

Answer:

Exponential functions!

Step-by-step explanation:

I think what you are dealing with is an exponential function!

So, to solve this, we need to first understand the main form of an exponential function:

f(x)=a*b^{x}

Where a is the number we start off with, and b is the constant thing(forgot what it was called).

Now, 150 is our starting number, so it is a, and b is our constant at which the initial term is multiplied, which is 3!

Also, defining variables:

Let f(x)= # of Bees

Let x= # of years

Thus, here is our equation:

f(x)=150*3^{x}

So, that's it!

Now, some of my answers have been recently deleted because they violated community guidelines for recommending something that shall not be named, so if you do really need help on the subject, look online! It is the largest source of information humanity has created! (also I cannot explain Exponential Functions in a single paragraph)

Stay Safe!

4 0
3 years ago
A telephone booth that is 8 ft tall casts a shadow that is 4ft long. Find the height of a lawn ornanment that casts a 2 ft shado
sukhopar [10]
8 ft /4 ft of shadow=x / 2 ft of shadow
x=(8 ft * 2 ft of shadow) / 4 ft of shadow=4 ft

height of a lawn ornament=4 ft
5 0
3 years ago
How to graph (x-2)(y-3)=0
kolbaska11 [484]

Step-by-step explanation:

here you want to isolate the y.

so what you do in the beginning is divide both sides by x-2.

then, you'll have y-3 = 0

so y = 3

this is a horizontal graph that crosses the y axis and y = 3.

5 0
2 years ago
Which pair of sneakers has the lowest sale price?
galben [10]

Answer:

15% off $30

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
A random sample of 16 students selected from the student body of a large university had an average age of 25 years. We want to d
kenny6666 [7]

Answer:

z=\frac{25-24}{\frac{2}{\sqrt{16}}}=2    

p_v =2*P(Z>2)=0.0455  

If we compare the p value and 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 conclude that the true mean differs from 24 at 5% of significance

Step-by-step explanation:

Data given and notation  

\bar X=25 represent the sample mean

\sigma=2 represent the sample population deviation for the sample  

n=16 sample size  

\mu_o =24 represent the value that we want to test

\alpha=0.05 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 24, the system of hypothesis would be:  

Null hypothesis:\mu = 24  

Alternative hypothesis:\mu \neq 24  

If we analyze the size for the sample is < 30 but we know the population deviation so 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".  

Calculate the statistic

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

z=\frac{25-24}{\frac{2}{\sqrt{16}}}=2    

P-value

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

p_v =2*P(Z>2)=0.0455  

Conclusion  

If we compare the p value and 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 conclude that the true mean differs from 24 at 5% of significance

6 0
3 years ago
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