What types of triangles have at least two acute angles?

A survey of 500 farmers showed that of the farmers, 121 grew only wheat, 108 grew only corn, 92 grew only oats, 195 grew wh

kondaur [170]

Answer:

The farmers in regions II, IV, and VI had exactly two of the three Summing the numbers in these regions 25 + 15 + 10 we find the that 50 farmers grew exactly two of the three.

5 0

4 years ago

What would f(-5.2) equal if f(x)=[[x]]

ira [324]

Answer:

gfjhhkuk

Step-by-step explanation:

gkguklhyj,vghm

srry i just need points for my testt :((

8 0

4 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

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

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

I really need help on my algebra review

Ilya [14]

Answer:

the slope becomes 3 and the graph moves up 7 units. 2. The slope becomes 5 and the graph moves down 2 units.

]

5 0

3 years ago

You are downloading songs to your mp3 player the ratio of pop songs to rock songs is 5 to 4 you download 40 pop songs how many r

dmitriy555 [2]

The ratio of pop songs to rock songs you download is 5:4, and 40 are pop songs. You want to find out how many rock songs there are.
First, add the ratios together: 5+4=9. This means that 9 will be your denominator, so 5/9 songs are pop and 4/9 songs are rock.
Now you know that pop songs account for 5/9 of the songs, this means that 40 is 5/9 of the total songs.
To find out how much 4/9 is, you need to divide 40 by 5 to get 1/9 and times that number by 4 to get 4/9:
40÷5=8×4=32 songs are rock songs

7 0

3 years ago

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