All categories

3 years ago

Find the intersection between thelines 2x + 3y = 5 and 3x + 4y = 6

1 answer:

4 0

An intersection point, is a shared coordinate. If the equations share a coordinate we can say they are "equal". There are two ways (algebraically) to find this point. Substitution or linear combination. Because the coefficients are different, let's use linear combination

First stack the equations on top of each other.

2x + 3y = 5
3x + 4y = 6

Next let's manipulate one or both of the equations to get coefficients that are equal, but opposite. For this I'm going to turn the coefficient of x into 6 and -6 respectively by multiplying the first equation by 3 and the second by -2 to get

6x + 9y = 15
-6x -8y = -12

Now we can combine the equations with addition and the x's will cancel leaving a single variable (which we can solve)

y = 3

Now plug this back into either original equation to find the x coordinate.

2x + 3(3) = 5

2x + 9 = 5

2x = -4

x = -2.

So your intersection point is (-2,3)

You might be interested in

Fcffffghvfdfffffgggggggg

Sauron [17]

Answer:

Yes

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the simplified form of 4 x 6^2 ÷ 3 + 7?

disa [49]

72/5 or 14.4 hope this helped you

6 0

2 years ago

Read 2 more answers

In a group of a hundred and fifty students attending a youth workshop in mombasa, 125 of them are fluent in kiswahili, 135 in en

jek_recluse [69]

Answer:

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

Step-by-step explanation:

Step(i):-

Given total number of students n(T) = 150

Given 125 of them are fluent in Swahili

Let 'S' be the event of fluent in Swahili language

n(S) = 125

The probability that the fluent in Swahili language

$P(S) = \frac{n(S)}{n(T)} = \frac{125}{150} = 0.8333$

Let 'E' be the event of fluent in English language

n(E) = 135

The probability that the fluent in English language

$P(E) = \frac{n(E)}{n(T)} = \frac{135}{150} = 0.9$

n(E∩S) = 95

The probability that the fluent in English and Swahili

$P(SnE) = \frac{n(SnE)}{n(T)} = \frac{95}{150} = 0.633$

Step(ii):-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = P(S) + P(E) - P(S∩E)

= 0.833+0.9-0.633

= 1.1

Final answer:-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

8 0

2 years ago

A recent broadcast of a television show had a 10 share, meaning that among 6000 monitored households with TV sets in use, 10%

Kisachek [45]

Answer:

Null and alternative hypothesis

$H_0: \pi \geq0.25\\\\H_1: \pi$

Test statistic $z=-26.82$

P-value $P=0$

The null hypothesis is rejected.

It is concluded that the proportion of households tuned into the program is less than 25%. The claim of the advertiser is rigth and got statistical support.

Step-by-step explanation:

We have to perform a hypothesis test of a proportion.

The claim is that less than 25% were tuned into the program, so we will state this null and alternative hypothesis:

$H_0: \pi \geq0.25\\\\H_1: \pi$

The signifiance level is 0.01.

The sample has a proportion p=0.1 and sample size of n=6000.

The standard deviation of the proportion, needed to calculate the test statistic, is:

$\sigma_p=\sqrt{\frac{\pi(1-\pi)}{N} } =\sqrt{\frac{0.25(1-25)}{6000} } =0.0056$

The test statistic is calculated as:

$z=\frac{p-\pi+0.5/N}{\sigma_p} =\frac{0.1-0.25+0.5/6000}{0.0056}=\frac{-0.1499}{0.0056} =-26.82$

As this is a one-tailed test, the P-value is P(z<-26.82)=0. The P-value is smaller than the significance level (0.01), so the effect is significant.

Since the effect is significant, the null hypothesis is rejected. It is concluded that the proportion of households tuned into the program is less than 25%. The claim of the advertiser is rigth and got statistical support.

3 0

3 years ago

Using the formula for initial velocity h=-16t2+VT the height of any object can be found by the time, Annalee serves a volleyball

Zolol [24]

Answer:

evvvvvvvvv

Step-by-step explanation:

errrrrrrrrrrrrrrv

5 0

2 years ago