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defon
3 years ago
13

A group of ladies meet for an afternoon tea party. They bring all their cats. In all, there are 26 heads and 84 feet. How many l

adies and how many cats are in the room
Mathematics
2 answers:
SSSSS [86.1K]3 years ago
7 0
H=p+c  (heads equals the sum of people and cats) since there are 26 heads:

p+c=26, solve for c, subtract p from both sides

c=26-p, now for feet...

2p+4c=f, using c found above gives you:

2p+4(26-p)=f and we are told that there are 84 feet so:

2p+4(26-p)=84  perform indicated multiplication on left side

2p+104-4p=84, combine like terms on left side

-2p+104=84  subtract 104 from both sides

-2p=-20  divide both sides by -2

p=10, and since c=26-p

c=26-10

c=16

So there are 10 people and 16 cats in the room.


Komok [63]3 years ago
6 0




X= ladies       y = number of cats 

<span>Each lady has 2 feet and each cat has 4 feet </span>

<span><=> </span>

<span>x + y = 26 (1) => number of characters </span>
<span>2x + 4y = 84 (2) => number of feet </span>

<span><=> operation : (1) * 2 </span>

<span>2x + 2y = 52 (1') </span>
<span>2x + 4y = 88 (2') </span>

<span><=> operation : (2') - (1') </span>

<span>2y = 26 </span>

<span><=> </span>

<span>y = 20 and x = 32- 16 = 16</span>

<span>10 ladies and 16 cats</span>
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A population of 50 deer are introduced into a wildlife sanctuary. It is estimated that the sanctuary can sustain up to 500 deer.
rewona [7]

Answer:

p1 = 70 deer

p2 = 98 deer

Step-by-step explanation:

We use the population growth expression:

N(t)=N_0\,(1+r)^t

which in our case renders:

N(t)=N_0\,(1+r)^t\\N(t)=50\,(1+0.4) ^t\\N(t)=50\,(1.4) ^t

Therefore, after 1 year the population would be:

N(1)=50\,(1.4) ^1=70

giving a total of 70 deer

and after two years, it would be:

N(2)=50\,(1.4) ^2=98

8 0
3 years ago
In a simple random sample of 300 boards from this shipment, 12 fall outside these specifications. Calculate the lower confidence
Lyrx [107]

Answer:

The 95% confidence interval for the percentage of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

Step-by-step explanation:

In a random sample of 300 boards the number of boards that fall outside the specification is 12.

Compute the sample proportion of boards that fall outside the specification in this sample as follows:

\hat p =\frac{12}{300}=0.04

The (1 - <em>α</em>)% confidence interval for population proportion <em>p</em> is:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

The critical value of <em>z</em> for 95% confidence level is,

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use a <em>z</em>-table.

Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.04\pm1.96\sqrt{\frac{0.04(1-0.04)}{300}}\\=0.04\pm0.022\\=(0.018, 0.062)\\\approx(1.8\%, 6.2\%)

Thus, the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

6 0
3 years ago
Reagan used 1,297 sq ft of wrapping paper to wrap gifts for everyone in her class. If each gift used the same amount of wrapping
seraphim [82]
She used 54.04 sq ft of wrapping paper
5 0
3 years ago
Find the area of the polygon with vertices (1, 2) (1, -5) (5, -5) (5,2)
Svetradugi [14.3K]

Answer:

28 square units

Step-by-step explanation:

These vertices are written in the form of their x and y coordinates. Let

A=(1,2) ; B =(1,-5) ; C =(5, -5) ; D=(5,2)

The x points are 1, 1, 5, 5

The y points are 2,-5, -5, 2

The x extremes range from 1 to 5. The length  in-between is 5-1 = 4

The y extremes range from -5 to 2. The length  in-between is 2-(-5) = 7

Area of the polygon enclosed by the coordinates

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5 0
3 years ago
How do you solve this?
nalin [4]
So u have to add all the sides up.
which will give you 6 + 6√3 + 6 +6√3
because their opposite and parallel lengths and widths are equal
that gives you 12 + 12√3
so ans is 12m + 12√3m
7 0
3 years ago
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