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3 years ago

What is the quotient? Negative StartFraction 3 over 8 EndFraction divided by negative one-fourth

Mathematics

1 answer:

anyanavicka [17]3 years ago

6 0

Answer:

$\dfrac{3}{2}$

Step-by-step explanation:

We are given 2 fractions.

1st fraction is Negative StartFraction 3 over 8

i.e.

$-\dfrac{3}{8}$

2nd fraction is negative one-fourth

i.e.

$-\dfrac{1}{4}$

We have to find the quotient when 1st fraction is divided by 2nd fraction.

<u>Definition</u> of quotient is given as:

Quotient is the result obtained when one number is divided by other number.

$-\dfrac{3}{8} \div -\dfrac{1}{4} \text{ is to be calculated.}$

$\div$ is converted to $\times$ and the 2nd fraction is reversed i.e.

if the fraction is $\frac{p}{q}$ it becomes $\frac{q}{p}$ and the sign $\div$ is changed to $\times$ .

Solving above:

$-\dfrac{3}{8} \times -\dfrac{4}{1}\\ \text {Multiplication/Division of 2 negative number gives a positive number}\\\Rightarrow \dfrac{3}{2}$

So, the quotient is $\frac{3}{2}$ .

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What is the line of symmetry?

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Answer:

The line of symmetry can be defined as the axis or imaginary line that passes through the center of the shape or object and divides it into identical halves.

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3 years ago

Please help! Evaluate the following function.<br> f(x) = -x – 8 for f(-1)

Anton [14]

Answer:

x - 8

Step-by-step explanation:

Plug in -1 for x and you get x - 8

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3 years ago

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4 years ago

Read 2 more answers

A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of seco

Brut [27]

Answer:

$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

$df=n_{A}+n_{B}-2=26+24-2=48$

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

$p_v =P(t_{(48)}>1.921)=0.0303$

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 mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X_{A}=48250$ represent the mean of elementary teachers

$\bar X_{B}=45630$ represent the mean for secondary teachers

$s_{A}=3900$ represent the sample standard deviation for elementary teacher

$s_{B}=5530$ represent the sample standard deviation for secondary teachers

$n_{A}=26$ sample size selected

$n_{B}=24$ sample size selected

$\alpha=0.05$ 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 of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers, the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{B}\leq 0$

Alternative hypothesis: $\mu_{A}-\mu_{B}>0$

We don't know the population deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{A}-\bar X_{B})}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

P-value

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

$df=n_{A}+n_{B}-2=26+24-2=48$

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

$p_v =P(t_{(48)}>1.921)=0.0303$

Conclusion

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3 years ago

Find a power series representation of e^x sin(x)

liberstina [14]

The Taylor series is defined by:
$f(x) = \sum \frac{f^n (a)}{n!} (x-a)^n$

Let a = 0.
Then its just a matter of finding derivatives and determining how many terms is needed for the series.

Derivatives can be found using product rule:
$f^n (x) = e^x g^{n-1} (x) + e^x g^n (x) \\ g^0 (x) = sin x$

Do this successively to n = 6.
$f^1 (x) = e^x (sinx +cos x) \\ f^2 (x) = e^x (2cos x) \\ f^3 (x) = e^x(2cos x - 2sin x) \\ f^4 (x) = e^x (-4 sin x) \\ f^5 (x) = e^x(-4sinx -4cos x) \\ f^6(x) = e^x(-8cos x)$

Plug in x=0 and sub into taylor series:
$e^x sin x = x+x^2 +\frac{x^3}{3} -\frac{x^5}{30}-\frac{x^6}{90}...$

If more terms are needed simply continue the recursive derivative formula and add to taylor series.

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3 years ago