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vladimir1956 [14]
3 years ago
8

Which data value has the highest frequency? 116 316 38 58

Mathematics
2 answers:
lara31 [8.8K]3 years ago
8 0
1/16.  that is the value that has the most x's on the graph.
andrew11 [14]3 years ago
3 0

Answer:

1/16

Step-by-step explanation:

Each tic mark represents 1/16.  This is because it takes 4 of them to make 1/4, and 4/16 = 1/4.

The first tic mark has the highest frequency, with 4 marks; this means 1/16 had the highest frequency.

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Which statement is true?
Karo-lina-s [1.5K]

Answer:

b, c, d.

Step-by-step explanation:

6 0
3 years ago
If AC=3x, AB=15, and BC=2x-1, how could we solve for x? A. Set 3x=15 B. Set 3x=2x-1 C. Set 3x+15=2x-1 D. Set 3x=15+ (2x-1) E. No
k0ka [10]
Hmm this is a very challenging question good luck finding someone who can answer it lol
5 0
3 years ago
The sum of two integers is 33. The larger is 6 more than twice the smaller. Find the two
grandymaker [24]

Answer:

24 and 9

Step-by-step explanation:

Hi there!

Let x be equal to the larger integer.

Let y be equal to the smaller integer.

<u>1) Construct equations</u>

  1. x+y=33 (The sum of two integers is 33)
  2. x=6+2y (The larger is 6 more than twice the smaller)

<u>2) Solve for one of the integer</u>

Isolate x in the first equation

x+y=33\\x=33-y

Plug the first equation into the second

33-y=6+2y

Combine like terms

33-6=2y+y\\27=3y\\9=y

Therefore, the smaller integer is 9.

<u />

<u>3) Solve for the other integer</u>

x+y=33

Plug in y (9)

x+9=33\\x=33-9\\x=24

Therefore, the larger integer is 24.

I hope this helps!

4 0
3 years ago
Read 2 more answers
The systolic blood pressure of adults in the USA is nearly normally distributed with a mean of 122 and standard deviation of 22
marin [14]

Answer: 4.27% of adults in the USA have stage 2 high blood pressure.

Step-by-step explanation:

Let x be a random variable that denotes a person with high blood pressure .

Given: Average blood pressure: \mu=122

Standard deviation: \sigma=22

Someone qualifies as having Stage 2 high blood pressure if their systolic blood pressure is 160 or higher.

The probability that an adult in the USA have stage 2 high blood pressure:

P(x\geq160)=P(\dfrac{x-\mu}{\sigma}}\geq\dfrac{160-122}{22})\\\\=P(z\geq1.72)\ \ \ [z=\dfrac{x-\mu}{\sigma}]\\\\=1-P(z

Hence, 4.27% of adults in the USA have stage 2 high blood pressure.

6 0
3 years ago
In an article regarding interracial dating and marriage recently appeared in a newspaper. Of 1719 randomly selected adults, 311
Bingel [31]

Answer:

Step-by-step explanation:

Hello!

The parameter of interest in this exercise is the population proportion of Asians that would welcome a person of other races in their family. Using the race of the welcomed one as categorizer we can define 3 variables:

X₁: Number of Asians that would welcome a white person into their families.

X₂: Number of Asians that would welcome a Latino person into their families.

X₃: Number of Asians that would welcome a black person into their families.

Now since we are working with the population that identifies as "Asians" the sample size will be: n= 251

Since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the variable distribution to normal.

Z_{1-\alpha /2}= Z_{0.975}= 1.965

1. 95% CI for Asians that would welcome a white person.

If 79% would welcome a white person, then the expected value is:

E(X)= n*p= 251*0.79= 198.29

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.79*0.21=41.6409

√V(X)= 6.45

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

198.29±1.965*6.45

[185.62;210.96]

With a 95% confidence level, you'd expect that the interval [185.62; 210.96] contains the number of Asian people that would welcome a White person in their family.

2. 95% CI for Asians that would welcome a Latino person.

If 71% would welcome a Latino person, then the expected value is:

E(X)= n*p= 251*0.71= 178.21

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.71*0.29= 51.6809

√V(X)= 7.19

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

178.21±1.965*7.19

[164.08; 192.34]

With a 95% confidence level, you'd expect that the interval [164.08; 192.34] contains the number of Asian people that would welcome a Latino person in their family.

3. 95% CI for Asians that would welcome a Black person.

If 66% would welcome a Black person, then the expected value is:

E(X)= n*p= 251*0.66= 165.66

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.66*0.34= 56.3244

√V(X)= 7.50

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

165.66±1.965*7.50

[150.92; 180.40]

With a 95% confidence level, you'd expect that the interval [150.92; 180.40] contains the number of Asian people that would welcome a Black person in their family.

I hope it helps!

5 0
3 years ago
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