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1 year ago

Divided 80 by my number, and then added 13. The result was 93. What was my number?

Mathematics

1 answer:

igor_vitrenko [27]1 year ago

4 0

The number is 6400.

We have given that,

divided 80 by my number, and then added 13.

<h3>What is the expression?</h3>

An expression or mathematical expression is a finite combination of symbols that is well-formed according to rules that depend on the context.

Suppose the number is x

$\frac{x}{80}+13=93$

Subtract 13 from both sides

$\frac{x}{80}+13-13=93-13$

$\frac{x}{80}=80$

Multiply both sides by 80

$\\\frac{80x}{80}=80\cdot \:80$

x=6400

Therefore the number is 6400.

To learn more about the number visit:

brainly.com/question/251701

#SPJ1

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Let f(x)=2.912345x^2+3.131579x-0.099999

Rudiy27

A.) Integers are positive and negative counting numbers. So, in order to find the integer coefficients, round off the coefficients in the equation to the nearest whole number. The function for g(x) is:

g(x) = 3x²+3x

B.) Substitute x=4 to the two functions.

f(x) = 2.912345x²<span>+3.131579x-0.099999
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C.) The percentage error is equal to:

Percentage error = |g(4) - f(4)|/f(4) * 100
Percentage error = |60 - 59.023837|/59.023837 * 100
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D.) If x is a large number, for example x=10 or x=20, then g(x) would be an overestimate. This is because the value of x is raised to the power of 2. So, as the x increases, the corresponding function would increase exponentially. Even at x=4, g(x) is already an overestimate. What more for larger values of x? That means that the gap from the true answer f(x) would increase.

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2 years ago

Evaluate the expression when m=7 and n=26.<br><br> n-63/m

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3 years ago

Read 2 more answers

A random sample of n = 40 observations from a quantitative population produced a mean x = 2.2 and a standard deviation s = 0.29.

zmey [24]

Answer:

$t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18$

$df=n-1=40-1=39$

Since is a one side test the p value would be:

$p_v =P(t_{(39)}>2.18)=0.0177$

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 then the population mes seems to be higher than 2.1 at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X=2.2$ represent the sample mean

$s=0.29$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =2.1$ 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 mean is higher than 2,1, the system of hypothesis would be:

Null hypothesis: $\mu \leq 2.1$

Alternative hypothesis: $\mu > 2.1$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-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:

$t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=40-1=39$

Since is a one side test the p value would be:

$p_v =P(t_{(39)}>2.18)=0.0177$

Conclusion

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