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3 years ago

14

A kennel in Akron charges $9 a night for dogs and $6 a night for cats. Write an expression that can be used to find the total of

a dog and cat staying at the kennel for 6 nights. Then solve the expression.
PLSSS

2 answers:

77julia77 [94]3 years ago

6 0

Answer:

90

Step-by-step explanation: 9 + 6 = 15

So, then you would multiply 15 x 6 which equals 90

matrenka [14]3 years ago

5 0

You would add 9+6=15 than you would multiply 15•6= 90.

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If sin(y°) = cos(x°), which of the following statements is true?

Firdavs [7]

Answer:

<u>y = w and ΔABC ~ ΔCDE</u>

Step-by-step explanation:

Given sin(y°) = cos(x°)

So, ∠y + ∠x = 90° ⇒(1)

And as shown at the graph:

ΔABC is aright triangle at B

So, ∠y + ∠z = 90° ⇒(2)

From (1) and (2)

<u>∴ ∠x = ∠z </u>

ΔCDE is aright triangle at D

So, ∠x + ∠w = 90° ⇒(3)

From (1) and (3)

<u>∴ ∠y = ∠w</u>

So, for the triangles ΔABC and ΔCDE

∠A = ∠C ⇒ proved by ∠y = ∠w
∠B = ∠D ⇒ Given ∠B and ∠D are right angles.
∠C = ∠E ⇒ proved by ∠x = ∠z

So, from the previous ΔABC ~ ΔCDE by AAA postulate.

So, the answer is <u>y = w and ΔABC ~ ΔCDE</u>

4 0

2 years ago

Read 2 more answers

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)

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Step-by-step explanation:

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