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3 years ago

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There is a cat in front of 2 cats, a cat behind 2 cats, and a cat between 2 cats. What is the least number of cats there could b

e in the line-up. /explain

2 answers:

Rufina [12.5K]3 years ago

6 0

3 cats. The first cat is in front of the other 2, the last cat is behind the other 2, and the middle is in between the other 2.

Maurinko [17]3 years ago

6 0

Answer:

3 cats

Step-by-step explanation:

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Please anyone I’m desperate

Savatey [412]

Answer:

the answer is B) 12a+6b

Step-by-step explanation:

3+9=12 so 3a+9a=12a,

7-1=6 so 7b-(1)b=6b

3 0

2 years ago

Read 2 more answers

Jilly has 8 pens. Three of the pens are purple. She will choose two pens to place in her school bag. What is the probability tha

Finger [1]

Answer:

3/8 is the probability

Step-by-step explanation:

3 0

2 years ago

the ratio of pine trees to elm trees in a park is 7:11. How many pine trees are there in the park if there are 68 more elm trees

Hitman42 [59]

Answer:

<em>The number of </em><em>pine tree is 119</em><em> and the number of </em><em>elm tree is 187.</em>

Step-by-step explanation:

The ratio of pine trees to elm trees in a park is 7:11

Let us assume that the number of pine tree is 7x and the number of elm tree is 11x.

It is also given that, there are 68 more elm trees as compared to the pine trees.

i.e $11x=7x+68$

$\Rightarrow 11x-7x=68$

$\Rightarrow 4x=68$

$\Rightarrow x=17$

Hence, the number of pine tree = $7x=7\times 17=119$ and the number of elm tree = $11x=11\times 17=187$

8 0

2 years ago

The random variable X is exponentially distributed, where X represents the waiting time to be seated at a restaurant during the

erastova [34]

Answer:

The probability that the wait time is greater than 14 minutes is 0.4786.

Step-by-step explanation:

The random variable <em>X</em> is defined as the waiting time to be seated at a restaurant during the evening.

The average waiting time is, <em>β</em> = 19 minutes.

The random variable <em>X</em> follows an Exponential distribution with parameter $\lambda=\frac{1}{\beta}=\frac{1}{19}$ .

The probability distribution function of <em>X</em> is:

$f(x)=\lambda e^{-\lambda x};\ x=0,1,2,3...$

Compute the value of the event (<em>X</em> > 14) as follows:

$P(X>14)=\int\limits^{\infty}_{14} {\lambda e^{-\lambda x}} \, dx=\lambda \int\limits^{\infty}_{14} {e^{-\lambda x}} \, dx\\=\lambda |\frac{e^{-\lambda x}}{-\lambda}|^{\infty}_{14}=e^{-\frac{1}{19} \times14}-0\\=0.4786$

Thus, the probability that the wait time is greater than 14 minutes is 0.4786.

7 0

2 years ago

Rearrange w= 3(2a + b) - 4 to make a the subject

Angelina_Jolie [31]

Answer:

$\boxed{a = \frac{w}{ 6} - \frac{b}{2} + \frac{2}{3} }$

Step-by-step explanation:

$Solve \: for \: a: \\ = > w=3(2a+b)-4 \\ \\ w=3(2a+b)4 \: is \: equivalent \: to \: 3(2a+b)-4= w: \\ = > 3(2a+b) - 4=w \\ \\ Expand \: out \: terms \: of \: the \: left \: hand \: side: \\ = > (3 \times 2a) + (3 \times b) - 4 = w \\ = > 6a+3b - 4=w \\ \\ Subtract \: 3b - 4 \: from \: both \: sides: \\ = > 6a + 3b - 4 - (3b - 4)=w - (3b - 4) \\ = > 6a = w - 3b + 4 \\ \\ Divide \: both \: sides \: by \: 6: \\ = > \frac{ \cancel{6}a}{ \cancel{6}} = \frac{w - 3b + 4}{6} \\ = > a = \frac{w}{6} - \frac{3b}{6} + \frac{4}{6} \\ = > a = \frac{w}{ 6} - \frac{b}{2} + \frac{2}{3}$

8 0

2 years ago