334% of what number is 33,400

Answer:

63.25 not an integer

Step-by-step explanation:

HCF(a,b)*LCM(a,b)=ab

11*368=64*x

x=11*368/64

x=63.25 not an integer, one of the given numbers must be incorrect

but you may use this method to find it yourself

Answer:

[tex]\boxed{\ x=2\ }[/tex]

Step-by-step explanation:

3(10-2x)=18

<=>

10-2x=18/3=6

<=>

2x=10-6=4

<=>

x= 4/2=2

Answer:

Graph B

Step-by-step explanation:

Simplify.

y - 4 = -3x - 15       Distribute

y = -3x - 11             Add 4 on both sides

The y-intercept should be negative, and option B has a negative y-intercept.

The graph of the given function will be represented by graph B so the correct answer is option B.

A graph is the representation of the data on the vertical and horizontal coordinates so we can see the trend of the data.

The graph of the function is attached with the answer below.

Simplify.

y - 4 = -3x - 15       Distribute

y = -3x - 11             Add 4 on both sides

The y-intercept should be negative, and option B has a negative y-intercept.

Therefore the graph of the given function will be represented by graph B so the correct answer is option B.

To know more about graphs follow

https://brainly.com/question/4025726

SPJ5

Answer:

[tex]81\pi[/tex]

Step-by-step explanation:

[tex]Area = \pi * r^{2} \\\pi *9^{2} =81\pi[/tex]

Answer:

81 π

Step-by-step explanation:

formula is radius squared times pi or π so the answer would be 9x9=81 and you said to leave in terms of π so the answer is 81 π.

Answer:

a) CI = (113,637.5  ,  124,672.5)

b) CI = (112,581  ,  125,729)

c) CI = (110,501.4  ,  127,808.6)

Step-by-step explanation:

You have the following information:

[tex]\overline{x}[/tex]: mean annual household income = 119,155

σ: standard deviation = 30,000

n: sample = 80

The interval of confidence is given by the following expression:

[tex]\overline{x}\pm Z_{\alpha/s}(\frac{\sigma}{\sqrt{n}})[/tex]

Z_α/2: distribution density factor

where α and Z_α/2 are given by the range of the confidence interval.

a) For a 90% confidence interval you have:

α = 1 - 0.9 = 0.1

Z_0.1/2 = Z_0.05 = 1.645   (found in a table of normal distribution)

You replace in the equation (1) to obtain the confidence interval:

[tex]119,155\pm (1.645)(\frac{30,000}{\sqrt{80}})\\\\=119,155\pm5,517.5[/tex]

Then, the confidence interval is (119,155 + 5,517.5 , 119,155 - 5,517.5  )

= (113,637.5  ,  124,672.5)

b) For a 95% confidence interval you have:

α = 1 - 0.95 = 0.05

Z_0.05/2 = Z_0.025 = 1.96

[tex]119,155\pm (1.96)(\frac{30,000}{\sqrt{80}})\\\\=119,155\pm 6,574.0[/tex]

The confidence interval is (112,581  ,  125,729)

c) For a 99% confidence interval:

α = 1 - 0.99 = 0.01

Z_0.01/2 = Z_0.005 = 2.58

[tex]119,155\pm (2.58)(\frac{30,000}{\sqrt{80}})\\\\=119,155\pm 8,653.6[/tex]

The confidence interval is (110,501.4  ,  127,808.6)

d) When the confidence level increases the width of the confidence increases too. This can be noticed in the normal distribution, when the confidence level is higher, the area of the tails is reduced, and so, the confidence interval is higher.

Answer:

A. The causal connection is valid. Students who spend more time and effort studying will be able to memorize more​ information, so their test grades will be higher.

Step-by-step Explanation:

The causal connection between the test grades of students and the amount of time and effort spent the students spend in studying and preparing for the test appears to be valid. This is valid because students who spend more time and effort studying would most likely be able to memorize more information of which they are most likely to come by in the test they take. Invariably, they'd be able to easily recall what they've memorize and give the right answers to the questions they are asked in the test, and this definitely will earn them higher test grades.

Answer:

0.2946

Step-by-step explanation:

Number of tosses, n = 200

P(obtaining a 5), p = 1/6

q = 1 - p = 5/6

Normal approximation for binomial distribution

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = np

= 200 x 1/6

= 33.33

Standard deviation = √npq

= √(200(1/6)(5/6) )

= 5.27

P(at most 30 fives) = P(X ≤ 30)

= P(Z < (30.5 - 33.33)/5.27) (continuity correction of 0.5 is added to 30)

= P(Z < -0.54)

= 0.2946

Answer:

15%

Step-by-step explanation:

Is means equals and of means multiply

60 = P * 400

Divide each side by 400

60/400 = P

.15 = P

Change to percent form

15%  is the percent

Answer:

the answer to the question you've asked is 15

Answer:

The group that has greater value of relative dispersion is the smokers group, as the coefficient of variationof their data is bigger than the coefficient of variation of the non-smokers group data.

CV smokers: 0.387

CV non-smokers: 0.234

Step-by-step explanation:

We will calculate the relative dispersion of each data set with its coefficient of variation (ratio of the standard deviation to the arithmetic mean).

Then, first we calculate the mean and standard deviation for the smokers data:

Mean: 43.7

Standard deviation: 286.5

[tex]M_s=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M_s=\dfrac{1}{12}(69.3+56+22.1+47.6+53.2+. . .+13.8)\\\\\\M_s=\dfrac{524.4}{12}\\\\\\M_s=43.7\\\\\\s_s=\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M_s)^2\\\\\\s_s=\dfrac{1}{11}((69.3-43.7)^2+. . . +(13.8-43.7)^2)\\\\\\s_s=\dfrac{3152}{11}\\\\\\s_s=286.5\\\\\\[/tex]

The mean and standard deviation for the non-smokers is:

Mean: 30.3

Standard deviation: 50.9

[tex]M_n=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M_n=\dfrac{1}{15}(28.6+25.1+26.4+34.9+28.8+. . .+13.9)\\\\\\M_n=\dfrac{453.8}{15}\\\\\\M_n=30.3\\\\\\s_n=\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M_n)^2\\\\\\s_n=\dfrac{1}{14}((28.6-30.3)^2+. . . +(13.9-30.3)^2)\\\\\\s_n=\dfrac{713.3}{14}\\\\\\s_n=50.9\\\\\\[/tex]

Now, we can calculate the coefficient of variation:

CV smokers:

[tex]CV_s=\dfrac{s_s}{M_s}=\dfrac{16.9}{43.7}=0.387[/tex]

CV non-smokers:

[tex]CV_n=\dfrac{s_n}{M_n}=\dfrac{7.1}{30.3}=0.234[/tex]

Answer:

64

Step-by-step explanation:

If the mean is 15, the sum of 5 numbers is:

Minimum value for the first four numbers would be:

Then the fifth number is:

So the maximum difference is:

Answer:

1. x = 2    y = 1

2. x = 17   y = 29/3

Step-by-step explanation:

1. Use elimination

-4x+3y=-5

4x-5y=3

x's cancel, so add them

-2y = -2

y = 1

substitute

4x -5(1) = 3

4x - 5 = 3

4x = 8

x = 2

2. Use elimination  

2x+3y=36

10x-6y=12

Multiply top equasion by 5

10x+30y = 360

10x-6y = 12

x's cancel so subract

36y = 348

y = 29/3

Substitute

2x+3(2/3)=36

2x+2 = 36

2x = 34

x = 17

Answer:

[tex] P(X \leq 1)= P(X=0) +P(X=1) [/tex]

And using the probability mass function we can find the individual probabiities

[tex]P(X=0)=(15C0)(0.82)^0 (1-0.82)^{15-0}=6.75x10^{-12}[/tex]

[tex]P(X=1)=(15C1)(0.82)^1 (1-0.82)^{15-1}=4.61x10^{-10}[/tex]

And replacing we got:

[tex] P(X \leq 1)= P(X=0) +P(X=1)= 4.68x10^{-10}[/tex]

And for this case yes we can conclude that 1 a significantly low number of adults requiring eyesight​ correction in a sample of 15 since the probability obtained is very near to 0

Step-by-step explanation:

Let X the random variable of interest "number of adults who need correction", on this case we now that:

[tex]X \sim Binom(n=15, p=0.82)[/tex]

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

We want to find this probability:

[tex] P(X \leq 1)= P(X=0) +P(X=1) [/tex]

And using the probability mass function we can find the individual probabiities

[tex]P(X=0)=(15C0)(0.82)^0 (1-0.82)^{15-0}=6.75x10^{-12}[/tex]

[tex]P(X=1)=(15C1)(0.82)^1 (1-0.82)^{15-1}=4.61x10^{-10}[/tex]

And replacing we got:

[tex] P(X \leq 1)= P(X=0) +P(X=1)= 4.68x10^{-10}[/tex]

And for this case yes we can conclude that 1 a significantly low number of adults requiring eyesight​ correction in a sample of 15 since the probability obtained is very near to 0

Answer:

Step-by-step

The null and the alternative hypothesis can be define as  follows,

Null Hypothesis; There is no significance difference between the proportions of non participating athletes in 1999 and 2000

[tex]H_0:(p_1-p_2)\neq 0[/tex]

Alternative Hypothesis: The proportion of non participating  athletes in  2000 will be more than the proportion of non participating athletes in 1999

[tex]H_1:(p_1-p_2)<0[/tex]

The proportion of nonparticipating athletes in 1999 is given by

[tex]\hat p_1 = \frac{x_1}{n_1} \\\\=\frac{159}{830} =0.1916[/tex]

The proportion of nonparticipating athletes in 2000 is given by

[tex]\hat p_2 =\frac{x_2}{n_1} \\\\=\frac{133}{825} =0.1612[/tex]

The pooled proportion can be calculated using the following formula

[tex]\hat p = \frac{x_1+x_2}{n_1+n_2} \\\\=\frac{159+133}{830+825} =0.1764[/tex]

under the null hypothesis, the test statistics can be calculated as follows

[tex]Z=\frac{\hat p_1 - \hat p_2}{\sqrt{\hat p \hat q(\frac{1}{n_1}+\frac{1}{n_2} ) } }[/tex]

[tex]=\frac{0.1916-0.1612}{\sqrt{(0.1764)(0.8236)(\frac{1}{830} +\frac{1}{825} )} } \\\\=1.6257[/tex]

Determine the P-value using the following formula

P-value = Normdist(1.6257)

=0.947993

Here, it can be observed that the P-value is greater than the level of the significance,

Hence, the null hypothesis fails to be rejected

Therefore it can be concluded that there is insufficient evidence to support that the proportions of non participating athletes in 2000 will be more than the proportions of non participating athletes in 1999

Answer:

[tex] P(\bar X>2.8)[/tex]

We can use the z score formula given by:

[tex] z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And replacing we got:

[tex] z=\frac{2.8 -2.32}{\frac{1.24}{\sqrt{32}}}=2.190 [/tex]

And using the normal standard distribution and the complement rule we got:

[tex] P(z>2.190 )= 1-P(z<2.190) = 1-0.986=0.014[/tex]

Step-by-step explanation:

For this case w eknow the following parameters:

[tex] \mu = 2.32[/tex] represent the mean

[tex]\sigma =1.24[/tex] represent the deviation

n= 32 represent the sample sze selected

We want to find the following probability:

[tex] P(\bar X>2.8)[/tex]

We can use the z score formula given by:

[tex] z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And replacing we got:

[tex] z=\frac{2.8 -2.32}{\frac{1.24}{\sqrt{32}}}=2.190 [/tex]

And using the normal standard distribution and the complement rule we got:

[tex] P(z>2.190 )= 1-P(z<2.190) = 1-0.986=0.014[/tex]

Answer:

0.55% probability that the mean childhood asthma prevalence for the sample is greater than 2.8​%. This means that a sample having an asthma prevalence of greater than 2.8% is unusual event, that is, unlikely.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If X is more than two standard deviations from the mean, it is considered an unusual outcome.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

[tex]\mu = 2.32, \sigma = 1.24, n = 43, s = \frac{1.24}{\sqrt{43}} = 0.189[/tex]

What is the probability that the mean childhood asthma prevalence for the sample is greater than 2.8​%?

This is 1 subtracted by the pvalue of Z when X = 2.8. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{2.8 - 2.32}{0.189}[/tex]

[tex]Z = 2.54[/tex]

[tex]Z = 2.54[/tex] has a pvalue of 0.9945

1 - 0.9945 = 0.0055

0.55% probability that the mean childhood asthma prevalence for the sample is greater than 2.8​%. This means that a sample having an asthma prevalence of greater than 2.8% is unusual event, that is, unlikely.

Answer:

Height of this missing bar would be 1

Step-by-step explanation:

Since there is 1 and only 1 quantity between 80-99.

Answer:

1

There is 1 number that is between 80 and 99 which is 99 so there should be 1 bar.

Step-by-step explanation:

Answer:

Laniece had 60 magazines

Step-by-step explanation:

Given: In the  first week, Tamia sold 30% more than Laniece. In the second  week, Tamia sold 12 magazines, but Laniece did not sell any. Tamia sold 50% more than Laniece by the end of the second  week

To find: Number of magazines sold by Laniece

Solution:

Let number of magazines sold by Laniece in the first week be x.

Number of magazines sold by Tamia in the first week = [tex]x+\frac{30}{100} x=\frac{130x}{100} =\frac{13x}{10}[/tex]

Number of magazines sold by Tamia in the second week = 12

Total number of magazines sold by Tamia at the end of the second week = [tex]\frac{13x}{10}+12[/tex]

Total number of magazines sold by Laniece at the end of the second week = x

According to question,

[tex]\frac{13x}{10}+12=x+\frac{50x}{100}=x+\frac{x}{2}\\\frac{13x}{10}+12=\frac{3x}{2}\\\frac{3x}{2}-\frac{13x}{10} =12\\\frac{15x-13x}{10}=12\\\frac{2x}{10}=12\\\frac{x}{5}=12\\x=60[/tex]

Answer:

only allow 3 decimals

90.284 is the answer we removed all others except for 3

Answer:

12 2/5 hours

Step-by-step explanation:

[tex]1+1+1\frac{1}{5} +1\frac{1}{5} +1\frac{1}{5} +1\frac{3}{5} +1\frac{3}{5} +1\frac{4}{5} +1\frac{4}{5} =\\\\2+3\frac{3}{5} +3\frac{1}{5} +3\frac{3}{5} =\\\\11\frac{7}{5} =\\\\12\frac{2}{5}[/tex]

12 2/5 hours have been logged in all.

Answer:

B is the midpoint of line segment AC

Answer:

14, 32

Step-by-step explanation:

45,15,44,17,40,20,31,25

this is combination of 2 series:

In the first series we can see the pattern as:

In the second series we can see the pattern as:

The series will continue as:

Answer:

14, 32

Step-by-step explanation:

lol :D

Answer:

x≥-7

Step-by-step explanation:

4x+5≥-23

Subtract 5 from each side

4x+5-5≥-23-5

4x≥-28

Divide each side by 4

4x/4≥-28/4

x≥-7

Answer:

X≥-7

Step-by-step explanation:

Step 1: Subtract 5 from both sides.

4x+5-5≥-23-5

4x ≥-28

Step 2: Divide both sides by 4.

4x/4 ≥-28/4

X ≥-7

Answer:

17.8 in x 21.8 in

Step-by-step explanation:

Given w=width and l=length

w*l=390

l=w+4, therefore w*(w+4)=390

w^2+4w=390

w^2+4w-390=0

Quadratic equation, solve as such

w=-21.8 or 17.8

Solution can't be negative so w=17.8 in

l=w+4 so l=21.8

Answer: 270

Step-by-step explanation:

The notation [tex](\frac{f}{g})(5)[/tex] means to divide [tex]\frac{f(5)}{g(5)}[/tex]. Now that we know we have to divide, we can plug them into this equation.

[tex]\frac{7+4(5)}{\frac{1}{2(5)} }=\frac{27}{\frac{1}{10} }[/tex]. We know that dividing by a fraction means to multiply by its reciprocal, so we'll do that.

[tex]27*10=270[/tex]

Answer:

16

Step-by-step explanation:

The range is the difference between the highest and lowest number of a data set. 22-6=16

Answer:

Step-by-step explanation:

Look at the population statistics. Let's say it contains:

- data on the age groups available in the population

- data on the probability that a child in the population has inadequate calcium intake OR data that a child in the population does not have the deficiency. If you're given one of these, the other can be gotten by subtracting the probability value given from 1.

So let's say there are children from ages 5 to 15 in this population and the probability that a child in this population has the deficiency is 0.23 (not all the children in this population of 5-15 year olds may have the deficiency) while the probability that a child in this population does not have the deficiency is [1-0.23] = 0.77

So if you pick a child randomly from the population and he has this deficiency, what is the probability that he or she is between 11 and 13 years old?

From ages 5-15, ages 11, 12 and 13 are 3 ages. The total number of ages is 11 ages.

3÷11 = 0.2727

This is the probability that a child picked or selected at random from the population is 11, 12, or 13 years old.

0.2727 × 0.23 = 0.0627

This is the probability that a child picked at random is BOTH within the age bracket 11 to 13 AND has the deficiency!

Apply this.

Answer:

6 4 5 6 7

Step-by-step explanation:

The mode would be 5. The first 2 letters of MODE are MO, which can be an abbreviation for MOST OFTEN.

Answer:

[tex]\frac{1}{3}[/tex]

Step-by-step explanation:

So first calculate what fraction of the circumference the arc is.

[tex]\frac{20}{360}=\frac{1}{18}[/tex]

Now the circumference is 6, so one eighteenth of that is [tex]\frac{1}{3}[/tex]

Answer:

(7, -11)

Step-by-step explanation:

If the point is shifted 3 to the right and 2 down, you just have to add 3 to the x-coordinate and subtract 2 from the y-coordinate. 4+ 3 = 7 and -9 - 2 is -11. So, the new point will be (7, -11).

Answer:

(7, -11)

Step-by-step explanation:

The point is translated three units to the right, and 2 units down.

[tex](4,-9)=>(4+3,-9-2)=>(7,-11)[/tex]

Point " ' " should be (7,-11)