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4 years ago

H0w to subtract 3/5 - 2/9 =

1 answer:

7 0

So here first you have to have same denominators to subtract and add.
to make same denominator you have to multiply something to them which meets up to their lowest common multiple (LCM). so here the easiest thing is multiply each denominator to the other but remember you have to multiply them to the numerator too.
3×9/5×9-2×5/9×5
so it becomes
27/45-10/45
17/45
as you cant bring to its simplest form since the answer in a prime no. leave it like that.

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PLEASE HELP!!

sp2606 [1]

Standard form is just putting (in this case) variables in alphabetical order. First we simplify- 2x+4=4y would become 1/2x+1=y. This is already in standard form, as numbers w/o variables come at the end. simplifying the next one is more tricky. first you get a variable/number alone-
9-2x-2y=4x+3
9-2y=6x+3
6-2y=6x
1-1/3y=x
1=x+1/3y
3=x+y
x+y=3
sorry if I was wrong and not of any help, But I do believe this is correct.

5 0

4 years ago

Read 2 more answers

Mike deposited $850 into the bank in July. From July to December, the amount of money he deposited into the bank increases by 25

lana [24]

It should be $10,062.50 because if you multiply 850 and 25% (turn to decimal .25) then you add 850 again and you get 10,062.50.

3 0

4 years ago

Read 2 more answers

Students in a representative sample of 69 second-year students selected from a large university in England participated in a stu

Serhud [2]

Answer:

95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

Step-by-step explanation:

We are given that for the 69 second-year students in the study at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.89.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

P.Q. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean procrastination score = 41

s = sample standard deviation = 6.89

n = sample of students = 69

$\mu$ = population mean estimate

<em>Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.</em>

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.9973 < $t_6_8$ < 1.9973) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.9973 & 1.9973 with P = 2.5%}

P(-1.9973 < $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ < 1.9973) = 0.95

P( $-1.9973 \times{\frac{s}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for </u> $\mu$ =[ $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ , $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ]

= [ $41-1.9973 \times{\frac{6.89}{\sqrt{69} } }$ , $41+1.9973 \times{\frac{6.89}{\sqrt{69} } }$ ]

= [39.34 , 42.66]

Therefore, 95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

5 0

3 years ago

Evaluate 15 over k when k=3

scZoUnD [109]

Step-by-step explanation:5

5 0

4 years ago

Yesterday I drank 16 cups of coffee. Today I drank 82% more cups. How many cups of coffee did I drink today?

sashaice [31]

Answer: 29.12 cups of coffee

Step-by-step explanation:

16 cups of coffee x 0.82 = 13.12

13.12 + 16 = 29.12

3 0

3 years ago