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3 years ago

What is 5 1/2 - 1 2/3

Mathematics

1 answer:

user100 [1]3 years ago

3 0

Answer:

3 5 /6

Step-by-step explanation:

Conversion a mixed number 5 1 /2 to a improper fraction: 5 1/2 = 5 1/ 2

= 5 · 2 + 1 /2 = 10 + 1 /2 = 11 /2

To find new numerator:

a) Multiply the whole number 5 by the denominator 2. Whole number 5 equally 5 * 2 /2 = 10 /2

b) Add the answer from previous step 10 to the numerator 1. New numerator is 10 + 1 = 11

c) Write previous answer (new numerator 11) over the denominator 2.

Five and one half is eleven halfs

Conversion a mixed number 1 2 /3 to a improper fraction: 1 2/3 = 1 2/ 3 = 1 · 3 + 2 /3 = 3 + 2 /3 = 5 /3

To find new numerator:

a) Multiply the whole number 1 by the denominator 3. Whole number 1 equally 1 * 3 /3 = 3 /3

b) Add the answer from previous step 3 to the numerator 2. New numerator is 3 + 2 = 5

c) Write previous answer (new numerator 5) over the denominator 3.

One and two thirds is five thirds

Subtract: 11 /2 - 5 /3 = 11 · 3 /2 · 3 - 5 · 2 /3 · 2 = 33 /6 - 10 /6 = 33 - 10 /6 = 23 /6

Simplify : 23/6 divide =3 5/6

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GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a m

Harrizon [31]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg

(b) The value of z test statistics is 2.50.

(c) We conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

(d) The p-value is 0.0062.

Step-by-step explanation:

We are given that Green Beam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury. Assuming a known standard deviation of 0.18 mg.

Let $\mu$ = average mg of mercury in compact fluorescent bulbs.

So, Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is no more than 3.50 mg}

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean mg of mercury = 3.59

$\sigma$ = population standard deviation = 0.18 mg

n = sample of bulbs = 25

So, test statistics = $\frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }$

= 2.50

The value of z test statistics is 2.50.

Now, P-value of the test statistics is given by;

P-value = P(Z > 2.50) = 1 - P(Z $\leq$ 2.50)

= 1 - 0.9938 = 0.0062

Now, at 0.01 significance level the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

6 0

3 years ago

Use the compound interest formula A =P(1 + r) t and the given information to solve for r.

Dmitry [639]

Answer:

Rounding to nearest hundredths gives us r=0.06.

So r is about 6%.

Step-by-step explanation:

So we are given:

$A=P(1+r)^t$

where

$A=2300$

$P=1600$

$t=6$ .

$A=P(1+r)^t$

$2300=1600(1+r)^6$

Divide both sides by 1600:

$\frac{2300}{1600}=(1+r)^6$

Simplify:

$\frac{23}{16}=(1+r)^6$

Take the 6th root of both sides:

$\sqrt[6]{\frac{23}{16}}=1+r$

Subtract 1 on both sides:

$\sqrt[6]{\frac{23}{16}}-1=r$

So the exact solution is $r=\sqrt[6]{\frac{23}{16}}-1$

Most likely we are asked to round to a certain place value.

I'm going to put my value for r into my calculator.

r=0.062350864

Rounding to nearest hundredths gives us r=0.06.

8 0

3 years ago

Jamie has 105 pieces of candy leftover from Halloween. She would like to distribute them evenly to the 7 kids on her block. Writ

garri49 [273]

Answer:

105/7=15

Step-by-step explanation:

equations have equal signs

3 0

2 years ago

Determine the type and number of solutions of x2+2x+3=0

erastova [34]

Solve the equation for x by finding a, b, and c of the quadratic then applying the quadratic formula.

x = −1 ± i√2

There's no real solutions

7 0

3 years ago

A school conference room can seat a maximum of 83 people. The principal and two counselors need to meet with the school’s studen

aniked [119]

Answer:

40 students

Step-by-step explanation:

We do not count the counselors or the principal because they are not students, so we do 83 - 3 = 80, now it is just simple division, for every student there is a parent, so we divide 80 in half, 80 / 2 = 40 students :D

4 0

2 years ago