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3 years ago

I need answers for all! thank you will mark brainlist!! please help me i’m begging you <3

Mathematics

1 answer:

Darina [25.2K]3 years ago

6 0

Step-by-step explanation:

y = 2/5 x - 9/5

when x = -1,

the value of y = 2/5(-1) - 9/5

= -2/5 - 9/5 = -11/5

when x = 0

=> y = 2/5(0) -9/5

= 0-9/5 = -9/5

when x=1

=> y = 2/5(1) - 9/5

= 2/5 - 9/5 = 7/5

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A sample of students from an introductory psychology class were polled regarding the number of hours they spent studying for the

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Answer:

$8.68,13.16$

Step-by-step explanation:

Hint- First we have to calculate the mean and standard deviation of the sample and then applying formula for confidence interval we can get the values.

Mean of the sample is,

$\mu=\dfrac{\sum _{i=1}^{24}a_i}{24}=\dfrac{262}{24}=10.92$

Standard deviation of the sample is,

$\sigma =\sqrt{\dfrac{\sum _{i=1}^{24}\left(x_i-10.92\right)^2}{24-1}}=5.6$

The confidence interval will be,

$=\mu \pm Z\dfrac{\sigma}{\sqrt{n}}$

Here,

Z for 95% confidence interval is 1.96, and n is sample size which is 24.

Putting the values,

$=10.92 \pm 1.96\cdot \dfrac{5.6}{\sqrt{24}}$

$=10.92 \pm 2.24$

$=8.68,13.16$

Confidence interval is used to express the degree of uncertainty associated with a sample.

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3 years ago

Inverse functions always go in the same direction as the original function. true or false?

zavuch27 [327]

False , that’s why they are inverse

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4 years ago

Find an equation in standard form of the parabola passing through the points (1, -2), (2,-2), (3,-4)

Yuliya22 [10]

The standard form of a parabola is y=ax²+bx+c
use the three given points to find the three unknown constants a, b, and c:
-2=a+b+c............1
-2=4a+2b+c......... 2
-4=9a+3b+c...........3
equation 2 minus equation 1: 3a+b=0..........4
equation 3 minus equation 2: 5a+b=-2.........5
equation 5 minus equation 4: 2a=-2, so a=-1
plug a=-1 in equation 4: -3+b=0, so b=3
Plug a=-1, b=3 in equation 1: -2=-1+3+c, so c=-4
the parabola is y=-x²+3x-4

double check: when x=1, y=-1+3-4=-2
when x=2, y=-4+6-4=-2
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Yes.

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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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MakcuM [25]

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