All categories

4 years ago

2x+5y-10=0 -4x+3y=-8

1 answer:

5 0

X = 35/13 and y = 12/13. Plz plz plz give me brainiest answer I only need 4 more to rank up. If u do I will show my work.

You might be interested in

the width of a rectangle is half as long as the length. the rectangle has a area of 32 square feet.what are the length and width

kari74 [83]

The answer to this question would be length is 4 and the width is 8.
My explanation: Well, all i did was think of what two numbers go into 32 and those were the first 2 that came into my mind ;-}

7 0

3 years ago

How many fifieths would three theths equal

kari74 [83]

45 you need to fix the question for example what does theths mean

3 0

3 years ago

Express 0.125 as a percent.

fenix001 [56]

Answer:

12.5 %

Step-by-step explanation:

5 0

4 years ago

Read 2 more answers

What is the width of a rectangle with an area of 5/8in2 and a length of 10 inches

Tju [1.3M]

A=lw

5/8=10w

w=1/16

Hope this helps :)

3 0

3 years ago

A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The

adelina 88 [10]

Answer:

The 90% confidence level is $19.15< L < 20.85$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 64$

The mean age is $\= x = 20 \ years$

The standard deviation is $\sigma = 4 \ years$

Generally the degree of freedom for this data set is mathematically represented as

$df = n - 1$

substituting values

$df = 64 - 1$

$df = 63$

Given that the level of confidence is 90% the significance level is mathematically evaluated as

$\alpha = 100 - 90$

$\alpha =$ 10 %

$\alpha = 0.10$

Now $\frac{\alpha }{2} = \frac{0.10}{2} = 0.05$

Since we are considering a on tail experiment

The critical value for half of this significance level at the calculated degree of freedom is obtained from the critical value table as

$t_{df, \frac{ \alpha}{2} } = t_{63, 0.05 } = 1.669$

The margin for error is mathematically represented as

$MOE = t_{df , \frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

substituting values

$MOE = 1.699 * \frac{4 }{\sqrt{64} }$

$MOE = 0.85$

he 90% confidence interval for the true average age of all students in the university is evaluated as follows

$\= x - MOE < L < \= x + E$

substituting values

$20 - 0. 85 < L < 20 + 0.85$

$19.15< L < 20.85$

4 0

3 years ago