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Jet001 [13]
3 years ago
13

Of the original 56 signers of the declaration of independence four of them are present in North Carolina if you selected one sig

n randomly how likely is it that he represent in North Carolina
Mathematics
1 answer:
kari74 [83]3 years ago
8 0
The chance of you selecting one randomly is 4/56, which simplifies to 1/14.
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Simplify <br><br><br> 2−4÷2+2^3<br><br> Please help
mars1129 [50]
Divide then multiply then do left to right
8 0
3 years ago
If the odds against Deborahs winning first prize in a chess tournament are 1 to 11 what is the probability of the event that she
Lemur [1.5K]

Answer: \dfrac{11}{12}

Step-by-step explanation:

Given

The odds against winning in a chess tournament are 1 to 11.

Odds is defined as the ratio of the probability of occurrence to the non-occurrence of event.

\therefore \text{Probability that event will occur is P'=}\dfrac{1}{1+11}\\\\\Rightarrow P'=\dfrac{1}{12}

Probability of non-occurrence i.e. she wins the first prize is

\Rightarrow P=1-\dfrac{1}{12}\\\\\Rightarrow P=\dfrac{11}{12}

7 0
3 years ago
The temperature at 5 a.m. was −7.4°C. By 9 a.m., the temperature was −4.7°C. How much warmer was the temperature at 9 a.m.?
Nana76 [90]

Answer:

(

Step-by-step explanation:

(-7°C)-(-4°C)

-7°C + 4°C

-3°C

3 0
3 years ago
Please answer correctly !!!!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!!
Norma-Jean [14]

Answer:

15

Step-by-step explanation:

its the same as the other angle on there

6 0
3 years ago
Read 2 more answers
A quality control specialist at a pencil manufacturer pulls a random sample of 45 pencils from the assembly line. The pencils ha
amm1812

Answer:

The P-value you would use to test the claim that the population mean of  pencils produced in that factory have a mean length equal to 18.0 cm is 0.00736.

Step-by-step explanation:

We are given that a quality control specialist at a pencil manufacturer pulls a random sample of 45 pencils from the assembly line.

The pencils have a mean length of  17.9 cm. Given that the population standard deviation is 0.25 cm.

Let \mu = <u><em>population mean length of  pencils produced in that factory.</em></u>

So, Null Hypothesis, H_0 : \mu = 18.0 cm     {means that the population mean of  pencils produced in that factory have a mean length equal to 18.0 cm}

Alternate Hypothesis, H_A : \mu\neq 18.0 cm     {means that the population mean of  pencils produced in that factory have a mean length different from 18.0 cm}

The test statistics that will be used here is <u>One-sample z-test</u> statistics because 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 length of pencils = 17.9 cm

           \sigma = population standard deviation = 0.25 cm

           n = sample of pencils = 45

So, <u><em>the test statistics</em></u> =  \frac{17.9-18.0}{\frac{0.25}{\sqrt{45} } }  

                                    =  -2.68

The value of z-test statistics is -2.68.

<u>Now, the P-value of the test statistics is given by;</u>

         P-value = P(Z < -2.68) = 1 - P(Z \leq 2.68)

                      = 1- 0.99632 = 0.00368

For the two-tailed test, the P-value is calculated as = 2 \times 0.00368 = 0.00736.

7 0
3 years ago
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