13. Carla drew two acute non-overlapping
angles that share a ray and labeled them
ZJLK and Z KLM. The two angles have
different measures. Carla says
ZILM is
greater than a right angle.
An acute angle is open
less than a right angle.​

Answer:

There are 6 total possibilities, 3 red faces and 3 prime numbers however, 2 and 3 are prime numbers and they are red as well so total successful outcomes = 3 + 3 - 2 = 4. This means that the answer is 4 / 6 or 2 / 3.

Answer:

[tex]\frac{2}{3}[/tex]

Step-by-step explanation:

[tex]1 \: 2\:3[/tex] are red

[tex]5[/tex] is prime.

Total of [tex]4[/tex] possibilities out of [tex]6[/tex]

[tex]\frac{4}{6}[/tex]

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:

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:

a) 0.3085

b) 2574

c) 0.0125

Step-by-step explanation:

mean (μ) = 2600 kcal/day and a standard deviation (σ) = 50 kcal/day

a) The z score is given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

[tex]z=\frac{x-\mu}{\sigma}=\frac{2650-2600}{50}=1[/tex]

From the normal distribution table, P(x > 2650) = P(z > 1) = 1 - P(z < 1) = 1 - 0.8413 = 0.1587

b) A probability of 30% corresponds with a z score of -0.52

[tex]z=\frac{x-\mu}{\sigma}\\-0.52=\frac{x-2600}{50} \\x-2600=-26\\x=2600-26\\x=2574[/tex]

c) For a sampling distribution of sample mean, the standard deviation is [tex]\frac{\sigma}{\sqrt{n} }[/tex]

The z score is given by:

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

n = 20

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n} }}=\frac{2625-2600}{\frac{50}{\sqrt{20} }}=2.24[/tex]

From the normal distribution table, P(x > 2625) = P(z > 2.24) = 1 - P(z < 2.24) = 1 - 0.9875 = 0.0125

Answer:

x= -19

Step-by-step explanation:

2x-4x-9=29

-2x=29+9

x=38/-2

= -19

Answer:

[tex]x=2-\sqrt{42}[/tex] and [tex]x=2+\sqrt{42} \\[/tex]

Step-by-step explanation:

Solve using the quadratic formula, which is [tex]x=\frac{-b + \sqrt{b^{2}-4ac }}{2a}[/tex] and [tex]x=\frac{-b-\sqrt{b^{2}-4ac } }{2a}[/tex]

Answer:

  see below

Step-by-step explanation:

The first subtraction has a zero result (blue) from the thousands digit, so we know the dividend has 4 in that place. The 5 in the 1s place of the dividend is brought down to fill the space on the bottom line. 6 goes into that number 0 times, so the final quotient digit is 0.

  4,925 = 6×820 +5

or

  4,925 ÷ 6 = 820 r5

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:

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]\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:

$9400

Step-by-step explanation:

1.5x4=6%

100-6=94

0.94x10000=9400

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:

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:

[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:

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:

-1

Step-by-step explanation:

plug in y, subtract 12 from seven, divide -5 by 5

The values of x and y are -1 and 4 respectively.

Linear equations are equation involving one or more expressions including variables and constants and the variables are having no exponents or the exponent of the variable is 1.

Linear equations may include one or more variables.

Given are a system of linear equations.

5x + 3y = 7

y = 4

We already have the value of y as 4.

Substituting that value of y = 4 in the first equation 5x + 3y = 7, we get,

5x + (3 × 4) = 7

5x + 12 = 7

Subtracting both sides by 12, we get,

5x + 12 - 12 = 7 - 12

5x = -5

Dividing both sides by 5, we get,

5x / 5 = -5 / 5

x = -1

Hence the value of x is -1 and the value of y is 4.

To learn more about Linear Equations, click on the link :

https://brainly.com/question/29739212

#SPJ5

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:

only allow 3 decimals

90.284 is the answer we removed all others except for 3

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:

5

Step-by-step explanation:

Divide both sides by 3 to get

1:5

Multiply by 5

To get 5:25

5 nurses for 25 patients

Answer:

5

Step-by-step explanation:

3 x 5 = 15

n x 5 = 25

n = 5

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:

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:

[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:

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:

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

So I try to help

Step-by-step explanation:

I don't no sorrry

Answer:

the first one!!

Step-by-step explanation:

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:

Step-by-step explanation:

Area of rectangle = l × b

= 23 × 6

= 138 feet

hope this helps

plz mark it as brainliest!!!!!!!

Answer:

Janas rate is of 4 miles per day.

Step-by-step explanation:

Her rate is the number of miles she ran per day.

We can solve this using a rule of three.

In 7 days, she ran 28 miles. How many miles she ran a day, that is, in one day?

1 day - x miles

7 days - 28 miles

[tex]7x = 28[/tex]

[tex]x = \frac{28}{7}[/tex]

[tex]x = 4[/tex]

Janas rate is of 4 miles per day.

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)