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ahrayia [7]
1 year ago
15

There are 25 trees on the Jackson’s property. Twenty percent of the trees are oak trees. Which equation can be used to find the

number of oak trees on the property?
StartFraction 25 divided by 1 Over 20 divided by 1 EndFraction = StartFraction 25 Over 20 EndFraction
StartFraction 100 times 5 Over 20 times 5 EndFraction = StartFraction 500 Over 100 EndFraction
StartFraction 20 times 4 Over 25 times 4 EndFraction = StartFraction 80 Over 100 EndFraction
StartFraction 20 divided by 4 Over 100 divided by 4 EndFraction = StartFraction 5 Over 25 EndFraction
Mathematics
1 answer:
ahrayia [7]1 year ago
8 0

Answer:

StartFraction 25 divided by 1 Over 20 divided by 1 EndFraction = StartFraction 25 Over 20 EndFraction

Step-by-step explanation:

Number of trees = 25

Percentage of aok trees = 20%

To obtain the Number of oak trees :

(Number of trees ÷ 1) ÷ (percentage ÷1)

(25 / 1) ÷ (20 / 1) = (25 /1) * (20 / 1) = 25 / 20

StartFraction 25 divided by 1 Over 20 divided by 1 EndFraction = StartFraction 25 Over 20 EndFraction

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Suppose Paul kicks a soccer ball straight up into the air with an initial velocity of 96 feet per second. The function f(x) = -1
Ludmilka [50]
The zeros are the values of t for which f(t) = 0.
i.e. <span>-16t^2 + 96t = 0
16</span>t^2 - 96t = 16t(t - 6)
16t = 0 or t - 6 = 0
t = 0 or t = 6
Therefore, the zeros are 0, 6

The time taken for the ball to hit the ground is the value of t when f(t) = 0.
i.e. t = 6.
3 0
2 years ago
A new curing process developed for a certain type of cement results in a mean compressive strength of 5000 kilograms per square
Sedbober [7]

Answer:

\alpha =0.0668

Step-by-step explanation:

Data given and notation  

The info given by the problem is:

n=25 the random sample taken

\mu =5000 represent the population mean

\sigma =100 represent the population standard deviation

The critical region on this case is \bar X so then if the value of \bar X \geq 4970 we fail to reject the null hypothesis. In other case we reject the null hypothesis

Null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the true mean is 5000, the system of hypothesis would be:  

Null hypothesis:\mu = 5000  

Alternative hypothesis:\mu \neq 5000  

Let's define the random variable X ="The compressive strength".

We know from the Central Limit Theorem that the distribution for the sample mean is given by:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

Find the probability of committing a type I error when H0 is true.

The definition for type of error I is reject the null hypothesis when actually is true, and is defined as \alpha the significance level.

So we can define \alpha like this:

\alpha= P(Error I)= P(\bar X

And in order to find this probability we can use the Z score given by this formula:

Z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And the value for the probability of error I is givn by:

\alpha= P(\bar X

4 0
1 year ago
8,000 is blank as 800
Alecsey [184]

8,000 is 10 times as much as 800

3 0
2 years ago
Read 2 more answers
Slater begins saving for college tuition. He invests $1000 in an account that pays 2.5% simple interest. How many years before h
Brrunno [24]
The formula is Interest = principle times rate times time in years.
                        I=prt        

p=1000
r= 0.025
t=x

To find the amount of interest that is earned in a specific time frame, subtract the final amount of money by the principal. 1500-1000=500.

500 = 1000(0.025)x
500 = 25x   
x= 20 years
 
6 0
1 year ago
Which properties are present in a table that represents an exponential function in the form y-b* when b &gt; 1?
Oksana_A [137]

Answer:

<u>Properties that are present are </u>

Property I

Property IV

Step-by-step explanation:

The function given is  y=b^x  where b > 1

Let's take a function, for example,  y=2^x

Let's check the conditions:

I. As the x-values increase, the y-values increase.

Let's put some values:

y = 2 ^ 1

y = 2

and

y = 2 ^ 2

y = 4

So this is TRUE.

II. The point (1,0) exists in the table.

Let's put 1 into x and see if it gives us 0

y = 2 ^ 1

y = 2

So this is FALSE.

III. As the x-value increase, the y-value decrease.

We have already seen that as x increase, y also increase in part I.

So this is FALSE.

IV. as the x value decrease the y values decrease approaching a singular value.

THe exponential function of this form NEVER goes to 0 and is NEVER negative. So as x decreases, y also decrease and approached a value (that is 0) but never becomes 0.

This is TRUE.

Option I and Option IV are true.

7 0
2 years ago
Read 2 more answers
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