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2 years ago

Pleaseeee help I need to know now!!!

Mathematics

1 answer:

PtichkaEL [24]2 years ago

4 0

Answer:

C. Infinite solutions

Step-by-step explanation:

I'll use the equal value method and make the second equation equal 6 instead of 18, so I have to divide the entire equation by 3

(6x+15y=18)/3 = 2x +5y = 6

The second equation is the same thing as the first one, meaning no matter what value they will equal the same thing, also meaning there are infinite solutions

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Find the GCF.<br><br> 18c3 and 24c3

12345 [234]

Answer:

6c³

Step-by-step explanation:

18c³ = 2 × 3 × 3 × c × c × c

24c³ = 2 × 2 × 2 × 3 × c × c × c

Now we show the common factors in bold:

18c³ = 2 × 3 × 3 × c × c × c

24c³ = 2 × 2 × 2 × 3 × c × c × c

The common factor are:

2, 3, c, c, c

GCF = 2 × 3 × c × c × c = 6c³

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3 years ago

Let an integer be chosen at random from the integers 1 to 30 inclusive. Find the probability that the integer chosen is not even

Nataly [62]

Exactly half of these integers are odd (1, 3, 5, ..., 29) so the probability of selecting an odd number is $\dfrac12$ .

3 0

3 years ago

Read 2 more answers

Express the trig ratios as fractions in simplest terms.

Neporo4naja [7]

Answer:

See below

Step-by-step explanation:

$\cos M=\frac{adjacent}{hypotenuse}=\frac{56}{70}=\frac{4}{5}\\\\\sin L=\frac{opposite}{hypotenuse}=\frac{56}{70}=\frac{4}{5}$

Hence, both ratios are equal to each other

3 0

2 years ago

+(7p-8q+6pq)+(q-2p+pq)-(10pq-p-4q)

lilavasa [31]

Answer:

Step-by-step explanation:

3 0

3 years ago

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$

In point c)

$\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\$

In point d)

using standard normal variate

$x=20.17\\\\ z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17$

8 0

3 years ago