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3 years ago

8. The book fair had a sale where 6 books were $20.46. If you wanted to buy 7 books, how much money would you need?

Mathematics

2 answers:

Margaret [11]3 years ago

6 0

Answer:

$23.87

Step-by-step explanation:

should be around there

Ganezh [65]3 years ago

4 0

Answer:

23.87

Step-by-step explanation:

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Find the total amount given the original price and tax or tip rate. Round to the nearest hundredth if necessary.

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magazine provided results from a poll of adults who were asked to identify their favorite pie. Among the respondents, % chose

Alex73 [517]

Complete Question

A magazine provided results from a poll of 500 adults who were asked to identify their favorite pie. Among the 500 respondents, 12 % chose chocolate pie, and the margin of error was given as plus or minus 5 percentage points.What values do $\r p , \ \r q$ , n, E, and p represent? If the confidence level is 90%, what is the value of $\alpha$ ?

Answer:

$\r p$ is the sample proportion $\r p = 0.12$

$n$ is the sample size is $n = 500$

$E$ is the margin of error is $E = 0.05$

$\r q$ represents the proportion of those that did not chose chocolate pie i.e $\r q = 1- \r p$

$\alpha = 10\%$

Step-by-step explanation:

Here

$\r p$ is the sample proportion $\r p = 0.12$

$n$ is the sample size is $n = 500$

$\r q$ represents the proportion of those that did not chose chocolate pie i.e

$\r q = 1- \r p$

$\r q = 1- 0.12$

$\r q = 0.88$

$E$ is the margin of error is $E = 0.05$

Generally $\alpha$ is the level of significance and it value is mathematically evaluated as

$\alpha = ( 100 - C )\%$

Where $C$ is the confidence level which is given in this question as $C = 90 \%$

$\alpha = ( 100 - 90 )\%$

$\alpha = 10\%$

5 0

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

Please help me with 8, 9, and 10 thanks! Show work too

Harrizon [31]

In 8, 9, 10, all the triangles are similar, meaning corresponding sides have the same ratio.

8. hypotenuse / (short side) = 25/x = x/9
x² = 25*9 = 5²×3²
x = √(5²×3²) = 5*3 = 15

9. (long side) / (short side) = x/7 = 9/x
x² = 7*9
x = √(9*7) = 3√7

10. (short side) / (long side) = x/48 = 48/64
x = 48²/64 = 36

3 0

3 years ago