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1 year ago

A + b = c

Mathematics

1 answer:

ipn [44]1 year ago

4 0

Answer:

a=9

b=9

c=18

Step-by-step explanation:

c(a+b)+a+b=36

2a+2b=36

a+b=18

c=18

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One number is larger than another by 9. If the greater number is increased by 10, and the lesser number is tripled, the sum of t

kkurt [141]

Answer:

The larger number is -6, the smaller number is -15

Step-by-step explanation:

We have two numbers, a and b.

We know that one number is larger than another by 9.

Then we can write:

a = b + 9

then a is larger than b by 9 units.

If the greater number is increased by 10 (a + 10) and the lesser number is tripled (3*b), the sum of the two would be -41:

(a + 10) + 3*b = -41

So we got two equations:

a = b + 9

(a + 10) + 3*b = -41

This is a system of equations.

One way to solve this is first isolate one variable in one of the two equations:

But we can see that the variable "a" is already isolated in the first equation, so we have:

a = b + 9

now we can replace that in the other equation:

(a + 10) + 3*b = -41

(b + 9) + 10 + 3*b = -41

now we can solve this for b.

9 + b + 10 + 3b = -41

(9 + 10) + (3b + b) = -41

19 + 4b = -41

4b = -41 -19 = -60

b = -60/4 = -15

b = -15

then:

a = b + 9

a = -15 + 9 = -6

a = -6

7 0

2 years ago

Plzzz help me Asap I will give you brainliest. I promise

kari74 [83]

He need to earn 48
Inequality : 80-32=x

8 0

3 years ago

Merta reports that 74% of its trains are on time. A check of 60 randomly selected trains shows that 38 of them arrived on time.

kenny6666 [7]

Answer:

No, the on-time rate of 74% is not correct.

Solution:

As per the question:

Sample size, n = 60

The proportion of the population, P' = 74% = 0.74

q' = 1 - 0.74 = 0.26

We need to find the probability that out of 60 trains, 38 or lesser trains arrive on time.

Now,

The proportion of the given sample, p = $\frac{38}{60} = 0.634$

Therefore, the probability is given by:

$P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}]$

P $(p\leq 0.634) = P[z\leq -1.87188]$

P $(p\leq 0.634) = P[z\leq -1.87] = 0.0298$

Therefore, Probability of the 38 or lesser trains out of 60 trains to be on time is 0.0298 or 2.98 %

Thus the on-time rate of 74% is incorrect.

6 0

3 years ago

3. Find the value of a and b when<br>5+√6/5-√6 = a+b√6

mr Goodwill [35]

The answer for A is,

$a = 5 + \frac{ \sqrt{6} }{ \sqrt{5} } - \sqrt{6} - \sqrt{6b}$

And the answer for B is,

$b = - 1 + \frac{5 + \frac{ \sqrt{6} }{ \sqrt{5} } - a }{ \sqrt{6} }$

hope it helps.

sorry of its wrong.

4 0

3 years ago

A company claims that the mean weight per apple they ship is 120 grams with a standard deviation of 12 grams. Data generated fro

Veronika [31]

Answer:

There is no sufficient evidence to reject the company's claim at the significance level of 0.05

Step-by-step explanation:

Let $\mu$ be the true mean weight per apple the company ship. We want to test the next hypothesis

$H_{0}:\mu=120$ vs $H_{1}:\mu\neq 120$ (two-tailed test).

Because we have a large sample of size n = 49 apples randomly selected from a shipment, the test statistic is given by

$Z=\frac{\bar{X}-120}{\sigma/\sqrt{n}}$ which is normally distributed. The observed value is

$z_{0}=\frac{122.5-120}{12/\sqrt{49}}=1.4583$ . The rejection region for $\alpha = 0.05$ is given by RR = {z| z < -1.96 or z > 1.96} where the area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well. Because the observed value 1.4583 does not fall inside the rejection region RR, we fail to reject the null hypothesis.

5 0

3 years ago