Please help me quick !!!

Express the following as a fraction in its simplest form.

pychu [463]

Given:

Consider the given expression is:

$\dfrac{h+f}{3}+\dfrac{f+k}{2}+\dfrac{4h-k}{5}$

To find:

The simplified form of the given expression.

Solution:

We have,

$\dfrac{h+f}{3}+\dfrac{f+k}{2}+\dfrac{4h-k}{5}$

Taking LCM, we get

$=\dfrac{10(h+f)+15(f+k)+6(4h-k)}{30}$

$=\dfrac{10h+10f+15f+15k+24h-6k}{30}$

$=\dfrac{34h+25f+9k}{30}$

Therefore, the required simplified fraction for the given expression is $\dfrac{34h+25f+9k}{30}$ .

6 0

2 years ago

Instructions:Select the correct answer. What is the general form of the equation of the given circle with center A? x2 + y2 + 6x

forsale [732]

Its the third one hope this helps

8 0

2 years ago

Read 2 more answers

Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature

Vesnalui [34]

Answer:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation

$\bar X=21.5$ represent the mean for the temperatures

$s=1.5$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =22$ represent the value that we want to test

$\alpha=0.01$ 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 less than 22C, the system of hypothesis would be:

Null hypothesis: $\mu \geq 22$

Alternative hypothesis: $\mu < 22$

If we analyze the size for the sample is > 30 and 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:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

P-value

We can calculate the degrees of freedom like this:

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

Since is a one left tailed test the p value would be:

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

5 0

2 years ago

NEED HELP ASAP

balu736 [363]

Answer:

∠4 and ∠3

∠4 and ∠5

∠3 and ∠6

Step-by-step explanation:

A pair of angles is said to be supplementary when the 2 angles add up to give 180°.

Two right angles add up to give 180°. Also, if we have two angles on a straight line, their sum = 180°, according to the linear pair property.

Thus, since m< 3 is given as 90°, m<4 = 90°.

m<4 + m<3 = 180°.

Therefore, <4 and <3 are supplementary.

<4 and <5, <3 and <6 are both linear pairs, therefore they are also supplementary.

6 0

3 years ago

Use the Law of Sines to find the measure of angle J to the nearest degree.

rjkz [21]

$\bf \textit{Law of sines}
\\ \quad \\
\cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\\\\
-----------------------------\\\\
\cfrac{sin(J)}{9.1}=\cfrac{sin(97^o)}{11}\implies sin(J)=\cfrac{9.1\cdot sin(97^o)}{11}
\\\\\\
\textit{now taking }sin^{-1}\textit{ to both sides}
\\\\\\
sin^{-1}\left[ sin(J) \right]=sin^{-1}\left( \cfrac{9.1\cdot sin(97^o)}{11} \right)
\\\\\\
\measuredangle J=sin^{-1}\left( \cfrac{9.1\cdot sin(97^o)}{11} \right)$

8 0

3 years ago