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3 years ago

You worked these hours this week: 1 1/2, 2 3/4, 3 3/4, 4, 5 1/2 How many hours did you work?

Mathematics

2 answers:

Paladinen [302]3 years ago

7 0

Answer is = 17 1/2
16 1/2
1 1/2 + 5 1/2 = 7
2 3/4 + 3 3/4 = 6 2/4 (6 1/2)
7 + 6 1/2 + 4 = 17 1/2

I am Lyosha [343]3 years ago

6 0

The answer is 17 1/2. of you need to show work just put 1 1/2+ 2 3/4 + 3 3/4+ 4 + 5 1/2 = 17 1/2

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Identify the values a, b, and c is the first step in using the quadratic formula to find the solution(s) to a quadratic equation

zlopas [31]

The values are a = 7, b = -9, c = -18.

<u>Step-by-step explanation:</u>

The given quadratic equation is $7x^{2} - 9x = 18$

The general form of the quadratic equation is $ax^{2} + bx + c = 0$

where,

a is the coefficient of x².
b is the coefficient of x.
c is the constant term.

Now, you have to modify the given quadratic equation similar to the general form of quadratic equation.

So, bring the constant term 18 to the left side of the equation for equating it to zero.

⇒ $7x^{2} - 9x - 18 = 0$

Compare the above equation with general form $ax^{2} + bx + c = 0$

⇒ a = 7

⇒ b = -9

⇒ c = -18

Therefore, the values of a, b, and c are 7, -9 and -18.

7 0

3 years ago

Read 2 more answers

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 9t + 9 cot(t/2), [π/4, 7π/4]

agasfer [191]

Answer:

the absolute maximum value is 89.96 and

the absolute minimum value is 23.173

Step-by-step explanation:

Here we have cotangent given by the following relation;

$cot \theta =\frac{1 }{tan \theta}$ so that the expression becomes

f(t) = 9t +9/tan(t/2)

Therefore, to look for the point of local extremum, we differentiate, the expression as follows;

f'(t) = $\frac{\mathrm{d} \left (9t +9/tan(t/2) \right )}{\mathrm{d} t}$ = $\frac{9\cdot sin^{2}(t)-\left (9\cdot cos^{2}(t)-18\cdot cos(t)+9 \right )}{2\cdot cos^{2}(t)-4\cdot cos(t)+2}$

Equating to 0 and solving gives

$\frac{9\cdot sin^{2}(t)-\left (9\cdot cos^{2}(t)-18\cdot cos(t)+9 \right )}{2\cdot cos^{2}(t)-4\cdot cos(t)+2} = 0$

$t=\frac{4\pi n_1 +\pi }{2} ; t = \frac{4\pi n_2 -\pi }{2}$

Where n $_i$ is an integer hence when n₁ = 0 and n₂ = 1 we have t = π/4 and t = 3π/2 respectively

Or we have by chain rule

f'(t) = 9 -(9/2)csc²(t/2)

Equating to zero gives

9 -(9/2)csc²(t/2) = 0

csc²(t/2) = 2

csc(t/2) = ±√2

The solutions are, in quadrant 1, t/2 = π/4 such that t = π/2 or

in quadrant 2 we have t/2 = π - π/4 so that t = 3π/2

We then evaluate between the given closed interval to find the absolute maximum and absolute minimum as follows;

f(x) for x = π/4, π/2, 3π/2, 7π/2

f(π/4) = 9·π/4 +9/tan(π/8) = 28.7965

f(π/2) = 9·π/2 +9/tan(π/4) = 23.137

f(3π/2) = 9·3π/2 +9/tan(3·π/4) = 33.412

f(7π/2) = 9·7π/2 +9/tan(7π/4) = 89.96

Therefore the absolute maximum value = 89.96 and

the absolute minimum value = 23.173.

7 0

3 years ago

A simple random sample of electronic components wil be selected to test for the mean lifetime in hours. Assume that component li

Serjik [45]

Answer:

189 components must be sampled.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.005 = 0.995$ , so Z = 2.575.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

Assume that component lifetimes are normally distributed with population standard deviation of 16 hours.

This means that $\sigma = 16$

How many components must be sampled so that a 99% confidence interval will have margin of error of 3 hours?

n components must be sampled.

n is found when M = 3. So

$M = z\frac{\sigma}{\sqrt{n}}$

$3 = 2.575\frac{16}{\sqrt{n}}$

$3\sqrt{n} = 2.575*16$

$\sqrt{n} = \frac{2.575*16}{3}$

$(\sqrt{n})^2 = (\frac{2.575*16}{3})^2$

$n = 188.6$

Rounding up:

189 components must be sampled.

4 0

3 years ago

Alison knits 1/10 of scarf in 4/5 of an hour. what fraction of a scarf can alison knit in 1 hour? and how do u solve it?

pav-90 [236]

For this case we can make the following rule of three:
1/10 scarf ------> 4/5 hour
x ------------------> 1 hour
Clearing the value of x we have:
x = (1 / (4/5)) * (1/10)
Rewriting we have:
x = (5/4) * (1/10)
x = 5/40
x = 1/8
Answer:
A fraction of a scarf that alison can knit in 1 hour is:
x = 1/8

8 0

2 years ago

Read 2 more answers

A high school has 36 players on the football team. the summary of the players' weights is given in the box plot. what is the int

Komok [63]

The interquartile range of the players' weights = 48 pounds.

The boxplot attached to this question is missing. It was obtained online and is attached to this solution of the question.

It should be noted that the following is true for a boxplot.

A box plot gives a visual representation of the distribution of the data, showing where most values lie and those values that greatly differ from the rest, called outliers.

The elements of the box plot are described thus;

The bottom side of the box represents the first quartile, and the top side, the third quartile. Therefore, the width of the central box represents the inter-quartile range.

The horizontal line inside the box is the median.

The lines extending from the box reach out to the minimum and the maximum values in the data set, as long as these values are not outliers. The ends of the whiskers are marked by two shorter horizontal lines.

Variables in the dataset, higher than Q3+(1.5×IQR) or lower than Q1-(1.5×IQR) are considered outliers and are usually shown using dots above the top whisker or below the bottom whisker.

So, it is evident that for this question,

First quartile = 174 pounds

Third quartile = 222 pounds

Inter Quartile Range = (Third quartile) - (First quartile)

= 222 - 174

= 48 pounds.

To know more about "interquartile"

Refer this link:

brainly.com/question/16988652

#SPJ4

6 0

1 year ago