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3 years ago

7

The prices for different prices at a grocery store are shown in the table below which of the following spices has the least pric

e per pound

1 answer:

slavikrds [6]3 years ago

5 0

Where is the picture??? I can’t help without the full question and answer choices:/

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Find f(-7) if f(x) = 2x + 8.

Rom4ik [11]

f(x) = 2x + 8

f(-7) = 2(-7) + 8

f(x) = -14 + 8

f(x) = -6

5 0

3 years ago

Read 2 more answers

Dilate the the triangle, scale factor = 3 (positive 3, ignore the

HACTEHA [7]

Answer:

The rule of dilation is $P'(x,y) = 3\cdot P(x,y)$ .

The vertices of the dilated triangle are $A'(x,y) = (-21, -18)$ , $B'(x,y) = (-15,-10)$ and $C'(x,y) = (-3,-15)$ , respectively.

Step-by-step explanation:

From Linear Algebra, we define the dilation by the following definition:

$P'(x,y) = O(x,y) + k\cdot[P(x,y)-O(x,y)]$ (1)

Where:

$O(x,y)$ - Center of dilation, dimensionless.

$k$ - Scale factor, dimensionless.

$P(x,y)$ - Original point, dimensionless.

$P'(x,y)$ - Dilated point, dimensionless.

If we know that $O(x,y) = (0,0)$ , $k = 3$ , $A(x,y) = (-7,-6)$ , $B(x,y) = (-5,-2)$ and $C(x,y) =(-1,-5)$ , then dilated points of triangle ABC are, respectively:

$A'(x,y) = O(x,y) + k\cdot [A(x,y)-O(x,y)]$ (2)

$A'(x,y) = (0,0) + 3\cdot [(-7,-6)-(0,0)]$

$A'(x,y) = (-21, -18)$

$B'(x,y) = O(x,y) + k\cdot [B(x,y)-O(x,y)]$ (3)

$B'(x,y) = (0,0) + 3\cdot [(-5,-2)-(0,0)]$

$B'(x,y) = (-15,-10)$

$C'(x,y) = O(x,y) + k\cdot [C(x,y)-O(x,y)]$ (4)

$C'(x,y) = (0,0) +3\cdot [(-1,-5)-(0,0)]$

$C'(x,y) = (-3,-15)$

The rule of dilation is:

$P'(x,y) = 3\cdot P(x,y)$ (5)

The vertices of the dilated triangle are $A'(x,y) = (-21, -18)$ , $B'(x,y) = (-15,-10)$ and $C'(x,y) = (-3,-15)$ , respectively.

7 0

3 years ago

assume the age of night school students is normally distributed. A simple random sample of 24 night school students had an avera

aleksley [76]

Let
n = number of data
s = standard deviation (sample)
S = standard deviation (population)

The working equations is

$\frac{(n-1) s^{2} }{ x^{2}_{right} } \ \textless \ S^{2} \ \textless \ \frac{(n-1) s^{2} }{ x^{2}_{left} }$

To find $x^{2}_{right}$ , : (1 - 0.90)/2 = 0.05

To find $x^{2}_{left}$ , : 1 - 0.05 = 0.95

Degrees of freedom = n-1 = 24 - 1 = 23

This is shown in the figure attached. Since there is no row for df=23, we interpolate. Thus,

$x^{2}_{left} = 13.093$

$x^{2}_{right} = 35.17$

Substitute all values,

$\frac{(24-1) 5.6^{2} }{ 35.17} \ \textless \ S^{2} \ \textless \ \frac{(24-1) 5.6^{2} }{ 13.093} }$

Thus the answer is,

$20.51\ \textless \ S^{2} \ \textless \ 55.09$

5 0

3 years ago

Sally has a GPA of 3.35. What percentage of the students have a GPA above her GPA?

vichka [17]

I'm hoping it might be around the .65 percentile?

3 0

3 years ago

A and b have opposite directions, and they both have a magnitude of 7. What must be true about ū and 51?

valentinak56 [21]

Equal but not parallel

6 0

4 years ago

Read 2 more answers