All categories

2 years ago

Help me answer this please

Mathematics

1 answer:

Diano4ka-milaya [45]2 years ago

3 0

The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it's pretty easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations.

You might be interested in

I need help please help me.

Nadusha1986 [10]

Answer:

20°.

Step-by-step explanation:

According to both the diagram and the presented angle measure, m∠RPS + m∠QPR = m∠QPS.

(4x + 27) + (9x - 115) = 107

4x + 9x + 27 - 115 = 107

13x - 88 = 107

13x = 195

x = 15

Now that we have the value of x, we can find the m∠QPR.

9x - 115

= 9 * 15 - 115

= 135 - 115

= 20

So, m∠QPR is 20°.

Hope this helps!

4 0

2 years ago

My friend sets out walking at a speed of 3 miles per hour. I set out behind her 5 minutes later at 4 miles per hour.

MissTica

3 m/h=0.05m/m
.05•5=0.25 m/h I believe?

7 0

2 years ago

Someone please help me ASAP please anyone I need help now

irinina [24]

(2165-59)/3=
.............

8 0

2 years ago

The middle school is 2.72 km from marshes House and 1.52 km from Ryan’s house how much further does Martha live from the middle

Alex_Xolod [135]

1.20kilometers farther away from Ryan’s house

6 0

2 years ago

A ferry will safely accommodate 82 tons of passenger cars. Assume that the meanweight of a passenger car is 2 tons with standard

Shalnov [3]

Answer:

The probability that the maximum safe-weight will be exceeded is <u>0.0455 or 4.55%</u>.

Step-by-step explanation:

Given:

Maximum safe-weight of 37 cars = 82 tons

∴ Maximum safe-weight of 1 car (x) = 82 ÷ 37 = 2.22 tons (Unitary method)

Mean weight of 1 car (μ) = 2 tons

Standard deviation of 37 cars = 0.8 tons

So, standard deviation of 1 car is given as:

$\sigma=\frac{0.8}{\sqrt{37}}=0.13$

Probability that maximum safe-weight is exceeded, P(x > 2.22) = ?

The sample is normally distributed (Assume)

Now, let us determine the z-score of the mean weight.

The z-score is given as:

$z=\frac{x-\mu}{\sigma}\\\\z=\frac{2.22-2}{0.13}\\\\z=\frac{0.22}{0.13}=1.69$

Now, finding P(x > 2.22) is same as finding P(z > 1.69).

From the z-score table of normal distribution curve, the value of area under the curve for z < 1.69 is 0.9545.

But we need the area under the curve for z > 1.69.

So, we subtract from the total area. Total area is 1 or 100%.

So, $P(z > 1.69) = 1 - P(z < 1.69)$

$P(z>1.69)=1-0.9545=0.0455\ or\ 4.55\%$

Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.

8 0

2 years ago