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3 years ago

a plant normally sells $6.50. today all plants are discounted by 25%. What is the sale price of the plant ?

Mathematics

2 answers:

Paraphin [41]3 years ago

5 0

Answer:

100% -> 6.50

100 - 25 = 75

75% -> 6.50/100 x 75

= 4.875

≈ 4.88

Cloud [144]3 years ago

3 0

Answer:

$4.875 or if rounded $4.88

Step-by-step explanation:

Hope this helped

also u just have to subtract 6.50 by 25%

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Combining like terms with negative coefficients and distribution <br> -5(3x+10)-2(4x-5) ?

N76 [4]

Answer:

-23x-40

Step-by-step explanation:

-5(3x+10)-2(4x-5)
-15x-50-8x+10
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3 years ago

What is the rate $2.50 for 10 items, as a unit rate?

Natali [406]

divide 2.20 by 10

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3 years ago

Let L be the line with parametric equations x=5+t,y=6,z=−2−3t. Find the vector equation for a line that passes through the point

scZoUnD [109]

Answer:

The required equations are

$(-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0$ and

$(-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0$ .

Step-by-step explanation:

The given parametric equation of the line, $L$ , is $x=5+t, y=6, z=-2-3t,$ so, an arbitrary point on the line is $R(x,y,z)=R(5+t, 6, -2-3t)$

The vector equation of the line passing through the points $P(-5,7,-8)$ and $R(5+t, 6, -2-3t)$ is

$\vec P + \lambda \vec{(PR)}=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((5+t-(-5))\hat i + (6-7)\hat j +(-2-3t-8)\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+t)\hat i -\hat j +(6-3t)\hat k\right)=0\;\cdots (i)$

The given equation can also be written as

$\frac {x-5}{1}=\frac {v-6}{0}=\frac{z+2}{-3}=t \; \cdots (ii)$

The standard equation of the line passes through the point $P_0(x_0,y_0,z_0)$ and having direction $\vec v= a_1 \hat i +a_2 \hat j +a_3 \hat k$ is

$\frac {x-x_0}{a_1}=\frac {y-y_0}{a_2}=\frac{z-z_0}{a_3}=t \;\cdots (iii)$

Here, The value of the parameter, $t$ , of any point $R$ at a distance $d$ from the point, $P_0$ , can be determined by

$|t \vec v|=d\;\cdots (iv)$

Comparing equations $(ii)$ and $(iii)$

The line is passing through the point $P_0 (5,6,-2)$ having direction $\vec v=\hat i -3 \hat k$ .

Note that the point $Q(5,6,-2)$ is the same as $P_0$ obtained above.

Now, the value of the parameter, $t$ , for point $R$ at distance $d=3$ from the point $Q(5,6,-2)$ can be determined by equation $(iv)$ , we have

$|t(\hat i -3 \hat k)|=3$

$\Rightarrow t^2|(\hat i -3 \hat k)|^2=9$

$\Rightarrow 10t^2=9$

$\Rightarrow t^2=\frac {9}{10}$

$\Rightarrow t=\pm \frac {3}{\sqrt {10}}$

Put the value of $t$ in equation $(i)$ , the required equations are as follows:

For $t= \frac {3}{\sqrt {10}}$

$(-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +\left(6-3\times \frac {3}{\sqrt {10}})\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0$

and for $t= -\frac {3}{\sqrt {10}}$ ,

$(-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\left (-\frac {3}{\sqrt {10}}\right))\hat i -\hat j +(6-3\times \left(-\frac {3}{\sqrt {10}}\right)\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0$

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3 years ago

Add 5 2/7+ 19 1/3+ 14 1/8= . Simplify the answer and write as a reduced mixed number.

elena55 [62]

I think that it it is 38

5 0

3 years ago

From 19 measurements of a pressure controller, the mean operating pressure is 4.97 kPa with a standard deviation of 0.0461 kPa.

Semenov [28]

Answer:

$4.97-2.10\frac{0.0461}{\sqrt{19}}=4.95$

$4.97+2.10\frac{0.0461}{\sqrt{19}}=4.99$

So on this case the 95% confidence interval would be given by (4.95;4.99)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X =4.97$ represent the sample mean

$\mu$ population mean (variable of interest)

s=0.0461 represent the sample standard deviation

n=19 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=19-1=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.10$

Now we have everything in order to replace into formula (1):

$4.97-2.10\frac{0.0461}{\sqrt{19}}=4.95$

$4.97+2.10\frac{0.0461}{\sqrt{19}}=4.99$

So on this case the 95% confidence interval would be given by (4.95;4.99)

5 0

3 years ago