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andrew11 [14]
3 years ago
14

Which is the best estimate of the sum of 56.78+5.06

Mathematics
1 answer:
faltersainse [42]3 years ago
7 0
61.84 is the answer. I hate how this has to be 20 letters
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Perimeter and Circumference! Only 5 questions
Wittaler [7]

Answer:

1) 17.5

2)43.96

3)20

4) 15.7

5)25

Step-by-step explanation:

ohhhh yes

4 0
3 years ago
Read 2 more answers
The question is in the picture
SVEN [57.7K]
The question asking for the circumference of the wheel. The formula is c= pie times diameter . C= 3.14* 26
you get c equals 81.64 inches.
6 0
3 years ago
Please help h get brainleist and points
Oksanka [162]

In order:

72 x 8 = 576

47 x 7 = 329

(7 x 3) x 7

21 x 7 = 147

5 0
3 years ago
wo balls are chosen randomly from an um containing 8 white, 4 black,and 2 orange balls. Suppose that we win $2 for each black ba
umka21 [38]

Answer:

The probability distribution is shown below.

Step-by-step explanation:

The urn consists of 8 white (<em>W</em>), 4 black (<em>B</em>) and 2 orange (<em>O</em>) balls.

The winning and losing criteria are:

  • Win $2 for each black ball selected.
  • Lose $1 for each white ball selected.

There are 8 + 4 + 2 = 14 balls in the urn.

The number of ways to select two balls is, {14\choose 2}=91 ways.

The distribution of amount won or lost is as follows:

Outcomes: WW  WO  WB  BB  BO  OO

X:                 -2      -1      1      4     2      0

Compute the probability of selecting 2 white balls as follows:

The number of ways to select 2 white balls is, {8\choose 2}=28 ways.

The probability of WW is,

P(WW)=\frac{n(WW)}{N}=\frac{28}{91}=0.3077

Compute the probability of selecting 1 white ball and 1 orange ball as follows:

The number of ways to select 1 white ball and 1 orange ball is, {8\choose 1}\times {2\choose 1}=16 ways.

The probability of WO is,

P(WO)=\frac{n(WO)}{N}=\frac{16}{91}=0.1758

Compute the probability of selecting 1 white ball and 1 black ball as follows:

The number of ways to select 1 white ball and 1 black ball is, {8\choose 1}\times {4\choose 1}=32 ways.

The probability of WB is,

P(WB)=\frac{n(WB)}{N}=\frac{32}{91}=0.3516

Compute the probability of selecting 2 black balls as follows:

The number of ways to select 2 black balls is, {4\choose 2}=6 ways.

The probability of BB is,

P(BB)=\frac{n(BB)}{N}=\frac{6}{91}=0.0659

Compute the probability of selecting 1 black ball and 1 orange ball as follows:

The number of ways to select 1 black ball and 1 orange ball is, {4\choose 1}\times {2\choose 1}=8 ways.

The probability of BO is,

P(BO)=\frac{n(BO)}{N}=\frac{8}{91}=0.0879

Compute the probability of selecting 2 orange balls as follows:

The number of ways to select 2 orange balls is, {2\choose 2}=1 ways.

The probability of OO is,

P(OO)=\frac{n(OO)}{N}=\frac{1}{91}=0.0110

The probability distribution of <em>X</em> is:

Outcomes:    WW     WO        WB         BB        BO         OO

X:                    -2          -1            1            4            2            0

P (X):           0.3077  0.1758  0.3516  0.0659  0.0879  0.0110

3 0
4 years ago
Sacramento County high school seniors have an average SAT score of 1,020. From a random sample of 144 Sacramento High School stu
kari74 [83]

Answer:

We need to conduct a hypothesis in order to check if the high school students are representative of the overall population (or if the true mean is 1020 or not), the system of hypothesis would be:    

Null hypothesis:\mu = 1020    

Alternative hypothesis:\mu \neq 1020  

Step-by-step explanation:

Previous concepts  and data given  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X=1100 represent the sample mean    

s=144 represent the sample standard deviation  

n=144 represent the sample selected  

\alpha significance level    

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the high school students are representative of the overall population (or if the true mean is 1020 or not), the system of hypothesis would be:    

Null hypothesis:\mu = 1020    

Alternative hypothesis:\mu \neq 1020    

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic  

We can replace in formula (1) the info given like this:    

t=\frac{1100-1020}{\frac{144}{\sqrt{144}}}=6.67      

P-value  

First we need to calculate the degrees of freedom given by:

df=n-1=144-1 =143

Then since is a two sided test the p value would be:    

p_v =2*P(t_{144}>6.67)=2.59x10^{-10}    

Conclusion    

If we compare the p value and a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is significant different from 1020 at 5% of significance.    

6 0
3 years ago
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