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vaieri [72.5K]
3 years ago
10

An average sized gray whale eats 9,849 pounds of plankton per week. On average,how much does it eat per day.Please show your wor

k.This is a division problem
Mathematics
2 answers:
Andrews [41]3 years ago
6 0
       1407
     _____
  7 | 9849
       7
     ------
       28
       28
     ------
         049
           49
      --------
             0
nata0808 [166]3 years ago
3 0
1,407 on average a day. 9,849/7=1,407
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According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 in
Zolol [24]

Answer:

The probability that the person is between 65 and 69 inches is 0.5403

Step-by-step explanation:

Mean height = \mu = 66

Standard deviation = \sigma = 2.5

We are supposed to find What is the probability that the person is between 65 and 69 inches i.e.P(65<x<69)

Z=\frac{x-\mu}{\sigma}

At x = 65

Z=\frac{65-66}{2.5}

Z=-0.4

Refer the z table for p value

P(x<65)=0.3446

At x = 69

Z=\frac{69-66}{2.5}

Z=1.2

P(x<69)=0.8849

So,P(65<x<69)=P(x<69)-P(x<65)=0.8849-0.3446=0.5403

Hence the probability that the person is between 65 and 69 inches is 0.5403

6 0
3 years ago
Given the function f(x)=6x-11, find f(-1/3).
Nadya [2.5K]

Answer:

-13

Step-by-step explanation:

(-1/3)*6 = -2

f(-1/3) = -2 -11

f(-1/3) = -13

7 0
3 years ago
While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem
Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = <em><u>population proportion</u></em>.

So, Null Hypothesis, H_0 : p = 0.013      {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, H_A : p > 0.013      {means that this is an unusually high number of faulty modems}

The test statistics that would be used here <u>One-sample z-test</u> for proportions;

                             T.S. =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~  N(0,1)

where, \hat p = sample proportion faulty modems= \frac{10}{367} = 0.027

           n = sample of modems = 367

So, <u><em>the test statistics</em></u>  =  \frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }

                                     =  2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0
3 years ago
Suppose that we are testing a coin to see if it is fair, so our hypotheses are: H0: p = 0.5 vs Ha: p ≠ 0.5. In each of (a) and (
aleksley [76]

Answer:

H_0: p = 0.5\\H_a: p \neq 0.5

a. We get 56 heads out of 100 tosses.

We will use one sample proportion test  

x = 56

n = 100

\widehat{p}=\frac{x}{n}

\widehat{p}=\frac{56}{100}

\widehat{p}=0.56

Formula of test statistic =\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}

                                       =\frac{0.56-0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}

                                       =1.2

refer the z table for p value

p value = 0.8849

a.  We get 560 heads out of 1000 tosses.

We will use one sample proportion test  

x = 560

n = 1000

\widehat{p}=\frac{x}{n}

\widehat{p}=\frac{560}{1000}

\widehat{p}=0.56

Formula of test statistic =\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}

                                       =\frac{0.56-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}

                                       =3.794

refer the z table for p value

p value = .000148

p value of part B is less than Part A because part B have 10 times the number the tosses.

6 0
2 years ago
Please help me understand this! Thanks a bunch if you help!
mariarad [96]
Segment BD equals 12<span />
6 0
3 years ago
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