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3 years ago

This is due soon... My math grade is bad and this is the last question please help! I'll give brainlist

Mathematics

1 answer:

klasskru [66]3 years ago

5 0

Answer:

12 + 2w > 40

w > 19

Minimum value of w is 20 feet.

Step-by-step explanation:

Consider the provided information.

Sharon has at most $25 to spend on her sister's birthday gift.

That means Sharon can spend not more than $25.

She already bought a knitting machine for her, which cost $14.99.

She would also like to get her sister some skeins of yarn to go with it. Each skein costs $2.75.

Let y represent the number of skeins of yarn.

Thus, the required inequality is:

Now solve the above inequality.

Skeins must be an integer.

Hence, the most number of skeins Sharon can afford to buy is 3.

ona wants to bake at most 30 loves of banana bread and nut bread for a bake sale. Each loaf of banana bread sells for $2.50 , And each loaf of nut bread sells for $2.75. Cora wants to make at least $44. Write a system of inequalities to model the situation

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1/6 divided by 6 by 6

Ahat [919]

the answer is 1/216

hope this might of helped :/

7 0

3 years ago

Read 2 more answers

If you are given the following stem and leaf display and asked to construct a frequency distribution chart, what would be the wi

Nuetrik [128]

Answer:

Width of intervals: 8

Step-by-step explanation:

We first look at how data is represented in a stem-leaf diagram.

Any number of the left (before -) is the stem and all numbers on right (after -) are the leaves. Each combination of stem and leaf represents one number. For example: 1 - 332 represents: 13, 13, 12.

Our data is as follows:

13, 13, 12, 24, 25, 31, 31, 35, 37, 42, 43, 41, 52, 51, 51, 52

To calculate the width of the frequency distribution chart, we have the following formula:

$Class\ width = \frac{Range}{Number\ of\ classes}$

The range of any data set = Maximum value in the data set - Minimum value in the data set

Maximum value in this case as seen from the data is 52 and minimum is 12.

Range = 52 - 12 = 40

Since we had only 5 stems in the data, we shall use that as the number of classes required in the frequency distribution chart.

$Class\ width = \frac{40}{5} = 8$

Hence, the class width in this data set will be 8.

To make the intervals, we begin from the minimum value and add 8 to it. The intervals will be:

12 - 20

20 - 28

28 - 36

36 - 44

44 - 52

Observe, that all the values of the stem lie within each interval.

For example, there are 3 values for stem 1: 12, 13, 13 and each lie in the first interval 12 - 20.

Next, the values of stem 2 are 24 and 25. Each of these value lie in the second interval 20 - 28; and henceforth.

8 0

4 years ago

A blind taste test will be conducted with 9 volunteers to determine whether people can taste a difference between bottled water

Aloiza [94]

Answer:

The answer is $\frac{1}{256}$

Step-by-step explanation:

If we assume that people cannot taste a difference between bottled water, then the probability of identifying tap water is 0.5

Thus, P(identify tap water)=0.5

The probability that at least 8 of the 9 people identify the tap water correctly is the sum of the probabilities

8 of 9 people identified correctly or
9 of 9 people identified correctly

Since P(identify tap water)=0.5 each probabilities are the same and equal to

$0.5^9$ = $\frac{1}{512}$

So we have $2*\frac{1}{512}$ = $\frac{1}{256}$

6 0

3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so