Answer:
32 years oldStep-by-step explanation:
3/4=24 so 1/4= 24÷3= 8
1/4=8
So to get 4/4 or Cameron's age it is 8×4=32yrs
[tex]answer \\ 32 \: years \: old \\ solution \\ mary's \: age = 24 \\ let \: cameron's \: age \: be \: x \\ given \\ \frac{3}{4} x = 24 \\ or \: x = 24 \times \frac{4}{3} \\ x = 32 \\ hope \: it \: helps[/tex]
Playbill magazine reported that the mean annual household income of its readers is $119,155 (Playbill, January 2006). Assume this estimate of the mean annual household in- come is based on a sample of 80 households, and based on past studies, the population standard deviation is known to be a = $30,000. a. Develop a 90% confidence interval estimate of the population mean. b. Develop a 95% confidence interval estimate of the population mean. c. Develop a 99% confidence interval estimate of the population mean. d. Discuss what happens to the width of the confidence interval as the confidence level is increased. Does this result seem reasonable? Explain.
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.
The mean percent of childhood asthma prevalence in 43 cities is 2.32%. A random sample of 32 of these cities is selected. What is the probability that the mean childhood asthma prevalence for the sample is greater than 2.8%? Interpret this probability. Assume that sigmaequals1.24%. The probability is nothing.
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.
60 is what percent of 400
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
Choose the following which is COMPLETELY correct:
Answer:
D
Step-by-step explanation:
Mean = (4+4+5+8+9) / 5
30 / 5
6
Median = put them in order and the one in the middle is the median.
4, 4, 5, 8, 9
Mode = the most common
4, 4, 5, 8, 9
What are the next two numbers in the pattern of numbers 45,15,44,17,40,20,31,25
Answer:
14, 32
Step-by-step explanation:
45,15,44,17,40,20,31,25
this is combination of 2 series:
45-44-40-31- ?15-17-20-25-?In the first series we can see the pattern as:
-1, -4, -9 = -1², -2², -3² so next difference must be -4², which is 31- 16= 14In the second series we can see the pattern as:
2, 3, 5 prime numbers, so next difference must be 7, which is 25+7=32The series will continue as:
45, 15, 44, 17, 40, 20, 31, 25, 14, 32Answer:
14, 32
Step-by-step explanation:
lol :D
If f(x)=7+4c and g(x) = 1/2x what is the value of (f/g)(5)
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]
Erythropoietin (EPO) is a banned drug used by athletes to increase the oxygen-carrying capacity of their blood. New tests for EPO were first introduced prior to the 2000 Olympic Games held in Sydney, Australia. Chance (Spring 2004) reported that of a sample of 830 world-class athletes, 159 did not compete in the 1999 World Championships (a year prior to the new EPO test). Similarly, 133 of 82:5 potential athletes did not compete in the 2000 Olympic Games. Was the new test effective in deterring an athlete's participation in the 2000 Olympics? If so, then the proportion of nonparticipating athletes in 2000 will be greater than the proportion of nonparticipating athletes in 1999, Use a 98% confidence interval to compare the two proportions and make the proper conclusion.
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
Please answer this correctly
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.
We wish to find the probability that a child from this population who has inadequate calcium intake is 11 to 13 years old. In other words, if you know that a child has inadequate calcium intake, what is the probability that the child is between 11 and 13 years old
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.
Find the probability that in 200 tosses of a fair die, we will obtain at most 30 fives
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
The HCF of two numbers is 11, and their L.C.M is 368. If one number is 64, then the other number is
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
Given that d is the midpoint of line segment ab and k is the midpoint of line segment bc, which statement must be true? (May give brainliest)
Answer:
B is the midpoint of line segment AC
The length of a rectangle is 4 inches longer than the width. If the area is 390 square inches, find the rectangle's dimensions. Round your answers to the nearest tenth of an inch.
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
Please answer this correctly
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:
Which of the following best forms the figure shown
Answer:
2 rays that meet at an endpoint
Step-by-step explanation: A ray starts with a dot, or point and continues on forever with an arrow. There are two rays in that drawing that start at the same endpoint.
Answer:
2 rays that meet at an endpoint.
Step-by-step explanation:
A ray is straight but has one endpoint and the other end go on infinitely.
A line is straight and goes on infinitely.
A line segment is straight and has two endpoints.
The picture shows two rays meeting at an endpoint.
Please help. I’ll mark you as brainliest if correct
Answer:
12 + -6i
a=12
b=-6
Step-by-step explanation:
( -4 + 3i ) ( -3 - 2i )
-4 * -3 = 12
3i * -2i= -6i
12 + -6i
Find the area of a circle with radius, r = 9cm.
Give your answer in terms of π .
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.
The lengths of nails produced in a factory are normally distributed with a mean of 5.02 centimeters and a standard deviation of 0.05 centimeters. Find the two lengths that separate the top 6% and the bottom 6%. These lengths could serve as limits used to identify which nails should be rejected. Round your answer to the nearest hundredth, if necessary.
Answer:
The length that separates the top 6% is 5.1 centimeters.
The length that separates the bottom 6% is 4.94 centimeters.
Step-by-step explanation:
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.
In this question, we have that:
[tex]\mu = 5.02, \sigma = 0.05[/tex]
Find the two lengths that separate the top 6% and the bottom 6%.
Top 6%:
The 100-6 = 94th percentile, which is X when Z has a pvalue of 0.94. So X when Z = 1.555.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.555 = \frac{X - 5.02}{0.05}[/tex]
[tex]X - 5.02 = 1.555*0.05[/tex]
[tex]X = 5.1[/tex]
So the length that separates the top 6% is 5.1 centimeters.
Bottom 6%:
The 6th percentile, which is X when Z has a pvalue of 0.06. So X when Z = -1.555.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.555 = \frac{X - 5.02}{0.05}[/tex]
[tex]X - 5.02 = -1.555*0.05[/tex]
[tex]X = 4.94[/tex]
The length that separates the bottom 6% is 4.94 centimeters.
Can you help me with this one don’t get it
Which graph represents this equation y-4= -3(x+5)
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.
What is a graph?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
A study of the effect of smoking on sleep patterns is conducted. The measure observed is the time, in minutes, that it takes to fall asleep. These data are obtained:
Smokers: 69.3 56.0 22.1 47.6 53.2 48.1 52.7 34.4 60.2 43.8 23.2 13.8
Non-Smokers: 28.6 25.1 26.4 34.9 28.8 28.4 38.5 30.2 30.6 31.8 41.6 21.1 36.0 37.9 13.9
Which group having greater value of relative dispersion and why?
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]
A survey showed that 82% of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If 15 adults are randomly selected, find the probability that no more than 1 of them need correction for their eyesight. Is 1 a significantly low number of adults requiring eyesight correction?
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
What is the solution set of the quadratic inequality x2+x-2
Answer:
The solution set is x = 1 and x = -2
Step-by-step explanation:
Solve 4x+5≥-23. show your work
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
Saved
250 mg
sing value in
50 mg
10 ml
X
Choice
Lesson 9: Problem Solving When the Percent Changes
Exit Ticket
Tamia and Laniece were selling magazines for a charity. In the
first week, Tamia sold 30% more than Laniece. In the second
week, Tamia sold 12 magazines, but Laniece did not sell any. If
Tamia sold 50% more than Laniece by the end of the second
week, how many magazines did Laniece sell? Choose any
model to solve the problem. Show your work to justify your
answer.
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]
Round 90.2844097979 to 3 decimals
Answer:
only allow 3 decimals
90.284 is the answer we removed all others except for 3
Find the range of the set of numbers shown by the box-and-whisker plot.
6
6
7
8
9
10
11
12
13
14
16
16
17
18
19
20
21
22
Answer:
16
Step-by-step explanation:
The range is the difference between the highest and lowest number of a data set. 22-6=16
1.solve for x 3(10 - 2x)=18
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
Determine whether the stated causal connection is valid. If the causal connection appears to be valid, provide an explanation. Test grades are affected by the amount of time and effort spent studying and preparing for the test. Choose the correct answer below
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.
b. The causal connection is valid. Students who spend more time and effort studying tend to be smarter, so their test grades are higher.
c. The causal connection is valid. When students spend more time and effort studying for a test, their test grades tend to be higher.
d. The causal connection is not valid.
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.
