How will the graph of the equations in a system appear if the system has no solution?

Jim bought 20 rosebushes last week. the regular price is $5.00 per bush, but last week they were on sale at 2 for $8.50. how muc

lorasvet [3.4K]

He saved $1.50 by buying them on discount

7 0

3 years ago

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A multiple choice question has 18 possible answers, only one of which is correct. Is it "significant" to answer a question cor

vodka [1.7K]

Answer: No

Step-by-step explanation:

Considering significant: p <= 0.05

A multiple choice question has 18 possible answers, only one of which is correct.

which means that the probability is 1 out 18, so

p = 1/18 = 0.056

As 0.056 > 0.050

Then, it is not significant

7 0

3 years ago

Two-fifths of the fish in Gary's fish tank are guppies.One-fourth of the guppies are red.What fraction of the fish in Gary's tan

DanielleElmas [232]

How many fish are there in all? We need that to answer the question.

7 0

3 years ago

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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

What is the predicted difference in sleep between individuals A and B, where individual A has 12 years of education and works 20

kari74 [83]

Answer: The predicted difference in sleep is 1251437 minutes.

Step-by-step explanation:

Since we have given that

Number of years of education = 12 years

Number of minutes a week = 2000

As we have

In case of individual A:

7 days = 2000 minutes

1 year = 365 days = $\dfrac{2000}{7}\times 365=104285$

12 years = $104285\times 12=1251420\ minutes$

In case of individual B:

7 days = 3000 minutes

1 year = 365 days = $\dfrac{3000}{7}\times 365=156428$

16 years = $156428\times 16=2502857$ minutes

So, the predicted difference in sleep between individuals A and B is given by

$2502857-1251420\\\\=1251437\ minutes$

Hence, the predicted difference in sleep is 1251437 minutes.

7 0

3 years ago