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marin [14]
3 years ago
11

Brianna used two congruent parallelograms and a square to create the figure shown.

Mathematics
2 answers:
koban [17]3 years ago
3 0

Answer: the awnser is 33 3/4

Step-by-step explanation: im literally doing the same test

Elis [28]3 years ago
3 0

Answer: 33 3/4 ft2

Step-by-step explanation:

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Assume each figure shown has the same orientation. Which figure is the image of square LMNP after a translation of (x, y) → (x +
creativ13 [48]

<u>Answer-</u>

Figure 4 is the image of square LMNP after a translation of (x, y) → (x + 5, y – 3)

<u>Solution-</u>

The co-ordinates of the vertices are,

L = (-3, 1)

M = (-1, 1)

N = (-1, -1)

P = (-3, -1)

A translation of (x, y) → (x + 5, y – 3) means, the point must be shifted 5 units right and then 3 units down.

After the transformation the new co-ordinates will be,

L = (-3+5, 1-3) = (2, -2)

M = (-1+5, 1-3) = (4, -2)

N = (-1+5, -1-3) = (4, -4)

P = (-3+5, -1-3) = (2, -4)

These are the co-ordinates of the figure no. 4

6 0
3 years ago
Read 2 more answers
7 × (–3) × (–2)^2 =
skad [1K]
<span>7 × (–3) × (–2)^2 
= </span>-21 × 4
= - 84

Hope it helps
5 0
3 years ago
Read 2 more answers
A blue die and a red die are thrown. B is the event that the blue comes up with a 6. E is the event that both dice come up even.
iVinArrow [24]

Answer:

Size of |E n B| = 2

Size of |B| = 1

Step-by-step explanation:

<em>I'll assume both die are 6 sides</em>

Given

Blue die and Red Die

Required

Sizes of sets

- |E\ n\ B|

- |B|

The question stated the following;

B = Event that blue die comes up with 6

E = Event that both dice come even

So first; we'll list out the sample space of both events

B = \{6\}

E = \{2,4,6\}

Calculating the size of |E n B|

|E n B| = \{2,4,6\}\ n\ \{6\}

|E n B| = \{2,4,6\}

<em>The size = 3 because it contains 3 possible outcomes</em>

Calculating the size of |B|

B = \{6\}

<em>The size = 1 because it contains 1 possible outcome</em>

8 0
3 years ago
Factor completely 6x²-4x -2
swat32
6x^2-4x -2 \\ \\ GCF = 2 \\ \\ 2( \frac{6x^2}{2} +  \frac{-4x}{2} -  \frac{2}{2} \ / \ factor \ out \ GCF \\ \\ 2(3x^2 - 2x - 1) \ / \ simplify \ terms \\ \\ 2(3x^2 + x - 3x - 1) \ / \ factor \\ \\ 2(x(3x+1) - (3x + 1)) \ / \ factor \ out \ common \ terms \\ \\ 2(3x+1)(x - 1) \ / \ factor \ out \ common \ term \\ \\ Answer: \fbox {2 (3x + 1)(x - 1)}
7 0
3 years ago
Read 2 more answers
The mean number of children among a sample of 15 low-income households is 2.8. The mean number of children among a sample of 19
gtnhenbr [62]

Answer:

Step-by-step explanation:

Let's put the data as below:

n1=15    x1=2.8      s1=1.6       and s1²=2.56

n2=19   x2=2.4     s2=1.7        and s2²=2.89

alpha= 0.05

To test the hypothesis:

H0= There is no sufficient evidence that low income household have fewer children

H1=There is sufficient evidence that low income household have fewer children

Assume that population variances are equal.

the t-static for two samples,

t=\frac{x1-x2}{Sp\sqrt{\frac{1}{n1}-\frac{1}{n2} \\} } } ~t with min (n1-1,n2-1)df

The pooled variance estimate Sp equals:

Sp^{2}=\frac{(n1-1)s1^{2}+(n2-1)s2^{2}  }{n1+n2-2}

Sp²=2.7456

Sp=1.65699

Degrees of freedom=n1+n2-2=32

Under null hypothesis:

tcal=\frac{|2.8 - 2.4|}{1.6599\sqrt{\frac{1}{15}+\frac{1}{19}  } }

tcal=0.6989

The critical value ttab=2.0369 for alpha=0.05

So we reject our null hypothesis H0

So there is sufficient evidence that low income households have fewer children than high income households

6 0
4 years ago
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