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Oliga [24]
3 years ago
5

How do you simplify numbers with exponents

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0
You write it out then you’ll get the exponent
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A family has two children. M represents male, and F represents female.
astraxan [27]

Answer:

Option c) Sample space=\{\bf FM, FF, MM, MF\}

Step-by-step explanation:

Given that a family has two children.

Let M represents male, and F represents female.

In probability theory, the sample space (also called sample description space or possibility space) of an experiment or random trial is the set of all possible outcomes or results of that experiment. A sample space is usually denoted using set notation, and the possible ordered outcomes are listed as elements in the set. It is common to refer to a sample space by the labels S, \Omega, or  \cup(for "universal set"). The elements of a sample space may be numbers, words, letters, or symbols. They can also be finite, countably infinite, or uncountably infinite.

Let S be the sample space of the given data.

Here the sample space is typically the set {male, female}, commonly written \{M, F\}

ie, S is the set \{male, female\}  or S=\{M, F\}

For the gender of the children,  corresponding sample space would be \{(male,male), (male,female), (female,male), (female,female)\} , commonly written as

\{MM,MF,FM,FF\}

S is the set \{(male,male), (male,female), (female,male), (female,female)\}  or  S=\{MM,MF,FM,FF\}

6 0
3 years ago
What is a formula used for determining the monetary predictor of neglect?
emmainna [20.7K]
<span>For the answer to the question above, I assume this is the complete question above,
What is a formula used for determining the monetary predictor of neglect
A) a2 + b2 = c2
B) B < PL
C) EA = π r2
D) X = Y/Z
If that's so, this is the answer
The answer is </span>C) EA = π r2
3 0
3 years ago
Pls answer dis question pls
Whitepunk [10]

Answer:

No solution

Step-by-step explanation:

4 0
3 years ago
Write the slope-intercept form of the equation of the line through the given point with the given slope.
RUDIKE [14]
Your answer is -1 lol
5 0
3 years ago
Read 2 more answers
The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of 10 batteries
Nat2105 [25]

Answer:

(1) We are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) A 95% lower confidence interval on mean battery life is 25.06 hr.

Step-by-step explanation:

We are given that a random sample of 10 batteries is subjected to an accelerated life test by running them continuously at an elevated temperature until failure, and the following lifetimes (in hours) are obtained: 25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                              P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~  t_n_-_1

where, \bar X = sample mean BAC = \frac{\sum X}{n} = 26 hr

             s = sample standard deviation = \sqrt{\frac{\sum(X - \bar X)^{2} }{n-1} } = 1.625 hr

             n = sample of batteries = 10

             \mu = population mean battery life

<em> Here for constructing a 95% lower confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

(1) It is stated that the manufacturer wants to be certain that the mean battery life exceeds 25 h.

Since we are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) So, a 95% lower confidence interval for the population mean, \mu is;

P(-1.833 < t_9) = 0.95  {As the lower critical value of t at 9

                                             degrees of  freedom is -1.833 with P = 5%}    

P(-1.833 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }) = 0.95

P( -1.833 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} ) = 0.95

P( \bar X-1.833 \times {\frac{s}{\sqrt{n} } } < \mu) = 0.95

<u>95% lower confidence interval for</u> \mu = [ \bar X-1.833 \times {\frac{s}{\sqrt{n} } } ]

                                                             = [ 26-1.833 \times {\frac{1.625}{\sqrt{10} } } ]

                                                            = [25.06 hr]

Therefore, a 95% lower confidence interval on mean battery life is 25.06 hr.

5 0
3 years ago
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