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2 years ago

The solution to (d+4)^2=9 is D=

Mathematics

2 answers:

Allushta [10]2 years ago

4 0

Answer:

$\boxed{\sf d = -1 \ \ \ or \ \ \ d = -7}$

Step-by-step explanation:

$\sf Solve \: for \: d \: over \: the \: real \: numbers: \\ \sf \implies {(d + 4)}^{2} = 9 \\ \\ \sf Take \: the \: square \: root \: of \: both \: sides: \\ \sf \implies \sqrt{ {(d + 4)}^{2} } = \sqrt{9} \\ \\ \sf 9 = {3}^{2} : \\ \sf \implies \sqrt{ {(d + 4)}^{2} } = \sqrt{ {(3)}^{2} } \\ \\ \sf \implies d + 4 = \pm 3 \\ \\ \sf \implies d + 4 = 3 \: \: \: \: \: or \: \: \: \: \: d + 4 = - 3 \\ \\ \sf Subtract \: 4 \: from \: both \: sides: \\ \sf \implies d + (4 - 4)= 3 - 4 \: \: \: \: \: or \: \: \: \: \: d + (4 - 4) = - 3 - 4 \\ \\ \sf \implies d = - 1 \: \: \: \: \: or \: \: \: \: \: d = - 7$

bonufazy [111]2 years ago

3 0

Answer
d = -1 or d = -7

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3 years ago

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

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3 years ago

Values for the labor force participation rate of women (LFPR) are published by the U.S. Bureau of Labor Statistics. A researche

Anna [14]

Answer:

(A) For this research we made use of the paired t test.

(B) For the t test, we conclude that in the null Hypothesis H₀, there was no difference between the average labor force participation rate for women between the years of 1968 and 1972 as against the alternate Hypothesis Hₐ.

From the output data we have, since the P value is 0.024 and is lesser than the alpha value of 0.05, we will accept the alternate Hypothesis instead of the null Hypothesis.

Note:Kindly find an attached copy of the complete question below.

Step-by-step explanation:

Solution

Recall that:

The U.S of Bureau of Labor Statistics have published Values for the labor force participation rate of women (LFPR).

We are more interested in if there was a difference between female participation 1968 and 1972, a rapid change in women.

so we will check values for 19 randomly selected cities for 1968 and 1972,

Now,

(a) Which of this test is most appropriate for this data

The test which will be more suitable for these data will be paired to test t.

Because the data is for labor force participation for two different years, we apply the t test. now that we are making a comparison of two different year times, we will apply the t test in this research.

In this case, the sample size for both the years evaluated should be same, since we need to pair each and all the data.

(b)By using the test you selected, explain your conclusion

Now, to test the data paired t test, we have already the null Hypothesis H₀: In this case there is no difference between the average labor force participation rate for women between the years of 1968 and 1972 as compared with the alternate Hypothesis Hₐ: there was a significant difference between the average labor force participation rate for women between the years of 1968 and 1972.

However, from the output, we have that the P value is 0.024 which is lower than the alpha value of 0.05.

In conclusion, the alternate Hypothesis accepted while the null Hypothesis is rejected.

Note: kindly find an attached copy of the complete question to this solution

below.