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3 years ago

The number of bagels sold daily for two bakeries is shown in the table:

Mathematics

2 answers:

nikitadnepr [17]3 years ago

6 0

Answer:

Step-by-step explanation:

den301095 [7]3 years ago

5 0

Answer: D

Reason: Both of the data shown is in no way symmetric

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When can you write ratios as a fraction?

k0ka [10]

Answer:

In a mathequation when needed

Step-by-step explanation:

4 0

2 years ago

A Lights-A-Lot quality inspector examines a sample of 25 strings of lights and finds that 6 are defective. What is the experimen

finlep [7]

Answer:

A 0.24 probability that one string is defective

Step-by-step explanation:

The sample probabiity:

$P=\frac{6}{25}$

$P=0.24$

If important to notice that this probability is from the sample, it can be extrapolated to the population but may not be the same.

5 0

3 years ago

Read 2 more answers

According to a certain website, wine critics generally use a wine-scoring scale to communicate their opinions on the relative qu

const2013 [10]

Answer:

$\bar x = 82$

$\sigma = 9.64$

Step-by-step explanation:

Given

$n = 12$

$87\ 91\ 86\ 82\ 72\ 91\ 60\ 77\ 80\ 79\ 83\ 96$

Solving (a); Point estimate of mean

To do this, we simply calculate the sample mean

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{87+ 91+ 86+ 82+72+91+60+77+80+79+83+96}{12}$

$\bar x = \frac{984}{12}$

$\bar x = 82$

Solving (b); Point estimate of standard deviation

To do this, we simply calculate the sample standard deviation

$\sigma = \sqrt{\frac{\sum(x-\bar x)^2}{n - 1}$

$\sigma = \sqrt\frac{(87-82)^2+ (91-82)^2+ ....+ (79-82)^2+ (83-82)^2+ (96-82)^2}{12-1}$

$\sigma = \sqrt\frac{1022}{11}$

$\sigma = \sqrt{92.91}$

$\sigma = 9.64$

<em>Note that: The sample mean and the sample standard deviation are the best point estimators for the mean and the standard deviation, respectively.</em>

<em>Hence, the need to solve for sample mean and sample standard deviation</em>

3 0

2 years ago

Which is equivalent to

Vanyuwa [196]

the answer is A couse divide all 5

8 0

2 years ago

First question, thanks. I believe there should be 3 answers

zysi [14]

Given: The following functions

$A)cos^2\theta=sin^2\theta-1$ $B)sin\theta=\frac{1}{csc\theta}$ $\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

$\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}$

$\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}$

$\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}$

$\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}$

$\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}$

Hence, the following are identities

$\begin{gathered} B)sin\theta=\frac{1}{csc\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

The marked are the trigonometric identities

3 0

1 year ago