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3 years ago

Group of kids rode horses to canyon. 20 eyes 30 legs between them. How many horses

1 answer:

8 0

If all were kids, there would be 20 legs. A horse has 2 more legs than a kid. The extra 10 legs can be attributed to there being 5 horses.

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Find the lengths of the missing side . Simplify all radicals !!!<br> help mee!!!!!!

larisa86 [58]

Answer:

$e = 13\sqrt{2}$

$f = 13\sqrt{2}$

Step-by-step explanation:

The ∆ given is an isosceles ∆ with a right angle measuring 90°, and two congruent angles measuring 45° each.

Using trigonometric ratio formula, we can find the lengths of the missing side as shown below:

Finding e:

$sin(\theta) = \frac{opp}{hyp}$

$sin(\theta) = sin(45) = \frac{\sqrt{2}}{2}$

hyp = 26

opp = e = ?

Plug in the values into the formula

$\frac{\sqrt{2}}{2} = \frac{e}{26}$

Multiply both sides by 26

$\frac{\sqrt{2}}{2}*26 = \frac{e}{26}*26$

$\frac{\sqrt{2}}{2}*26 = e$

$\frac{\sqrt{2}}{1}*13 = e$

$13\sqrt{2} = e$

$e = 13\sqrt{2}$

Since side e is of the same length with side f, therefore, the length of side f = $13\sqrt{2}$

3 0

3 years ago

Graph the circle x^2+y^2+2x+4y-44=0

Natasha_Volkova [10]

Answer:

See explanation

Step-by-step explanation:

First, convert this to standard form. By completing the square, you can do the following:

$(x^2+2x+1)-1+(y^2+4y+4)-4-44=0$

$(x+1)^2+(y+2)^2=49$

This is a circle with center (-1,-2), and radius 7. It looks like the one I graphed below. Hope this helps!

6 0

3 years ago

I NEED HELP ASAP IM DOING IT NOW. !!!!!!!!!!!!!!!!!!!!!!! Lily likes to collect records. Last year she had 12 records in her col

ratelena [41]

Answer:

12/10 =1.2 - 1 = 0.2 x 100 = 20%

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the hcf of 45 and 100

Ira Lisetskai [31]

Answer:

45 100

/ \ / \

3 × 15 2×50

/ \ / \

3× 5 2×25

/ \

5×5

45 : 3×3×5

100: 2×2×5×5

common : 3×5

HCF: 15

6 0

3 years ago

A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with t

julia-pushkina [17]

Answer:

a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.

b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.

c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Step-by-step explanation:

For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\lambda$ is the mean in the given interval.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The Poisson distribution can be approximated to the normal with $\mu = \lambda, \sigma = \sqrt{\lambda}$ , if $\lambda>10$ .

Poisson variable with the mean 3

This means that $\lambda= 3$ .

(a) At least 3 in a week.

This is $P(X \geq 3)$ . So

$P(X \geq 3) = 1 - P(X < 3)$

In which:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Then

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768$

0.5768 = 57.68% probability that the shop sells at least 3 in a week.

(b) At most 7 in a week.

This is:

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

In which

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

$P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680$

$P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008$

$P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504$

$P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216$

Then

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988$