The systolic blood pressure of adults in the USA is nearly normally distributed with a mean of 120 and standard deviation of 18 . Someone qualifies as having Stage 2 high blood pressure if their systolic blood pressure is 160 or higher. a. Around what percentage of adults in the USA have stage 2 high blood pressure

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Explanation:

We undo the "minus 6" by adding 6 to both sides.

Also, we undo the "+s" by subtracting s from both sides

-----------

So we have these steps

P = r+s-6

P+6 = r+s-6+6 .... adding 6 to both sides

P+6 = r+s

r+s = P+6

r+s-s = P+6-s ..... subtracting s from both sides

r = P + 6 - s

Answer:

P+6-s=r

Step-by-step explanation:

Hi there!

We are given the equation P=r+s-6 and we need to solve for r

To do that, we need to isolate r onto one side, and have everything else on the other.

Here is the equation:

P=r+s-6

start by adding 6 to both sides to clear it from the right side

P+6=r+s

now subtract s from both sides to clear it from the right side

P+6-s=r

now everything that isn't r is on the left side, and r is by itself on the right side. P+6-s=r is the answer.

Hope this helps!

Complete Question

Complete is Attached  Below

Answer:

Option D

Step-by-step explanation:

From the question we are told that:

Sample size [tex]n=26[/tex]

Student scoring [tex]65-100 n'=12[/tex]

Generally the equation for probability of a student having score between 65 and 100 is mathematically given by

 [tex]P(65-100)=\frac{12}{26}[/tex]

 [tex]P(65-100)=12/26[/tex]

 [tex]P(65-100)=0.462[/tex]

Option D

Answer:

The answer is "sometimes".

Step-by-step explanation:

A one-way ANOVA was merely performed on one collected data and the null hypothesis was rejected after an ANOVA F test. Assume we could randomize ANOVA block design with the same information. This null hypothesis for full equality is sometimes rejected for the randomized complete block design ANOVA. Therefore we understand the use of randomized ANOVA block if the null hypothesis is denied of a one-way ANOVA but the rejection of a null RBD ANOVA hypothesis isn't conditional mostly on denial of the yet another ANOVA null.

Answer:

10. CD + DE = CE

11. BC + CE = BE

Step-by-step explanation:

10. CD and DE lie on a straight line, therefore, CD + DE = CE based on the segment addition postulate.

11. BC and CE lie on a straight line, therefore, BC + CE = BE based on the segment addition postulate.

Answer:

the magnitude of a real number without regard to its sign.

Step-by-step explanation:

For example, |-3| would just be a 3 in general, no negative sign in the front.

hope this answers your confusion.

Answer:

A

Step-by-step explanation:

ITS OPTION (A)

PLZ MARK ME BRAINLIEST..

Hello,

P(A)=0.4

P(B)=0.3

P(AUB)+P(A∩B)=P(A)+P(B)

P(AUB)=0.4+0.3-0.1=0.6

Answer C

The correct answer is option (C).

P(A ∪ B) = 0.6

If A, B are two different events then P(A U B) = P(A) + P(B) - P(A ∩ B)

We have been given, P(A) = 0.4, P(B) = 0.3

From given Venn diagram,

P(A ∩ B) = 0.10

Now, P(A U B) = P(A) + P(B) - P(A ∩ B)

⇒ P(A U B) = 0.4 + 0.3 - 0.10

⇒ P(A ∪ B) = 0.6

Therefore, the correct answer is option (C) .6

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Answer:

[tex]\sigma = 121.53[/tex]

Step-by-step explanation:

Required

The population standard deviation

First, calculate the population mean

[tex]\mu = \frac{\sum x}{n}[/tex]

This gives:

[tex]\mu = \frac{148+ 329+ 491 +167+ 228+285+ 441}{7}[/tex]

[tex]\mu = \frac{2089}{7}[/tex]

[tex]\mu = 298.43[/tex]

The population standard deviation is:

[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}[/tex]

So, we have:

[tex]\sigma = \sqrt{\frac{(148 - 298.43)^2 + ..........+ (441- 298.43)^2}{7}}[/tex]

[tex]\sigma = \sqrt{\frac{103387.7143}{7}}[/tex]

[tex]\sigma = \sqrt{14769.6734714}[/tex]

[tex]\sigma = 121.53[/tex]

s hard and too long I'm only of class 13

Answer:

0.0106 = 1.06% probability that 2 or fewer will withdraw

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they withdraw, or they do not. The probability of an student withdrawing is independent of any other student, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

25% of its students withdraw without completing the introductory statistics course.

This means that [tex]p = 0.25[/tex]

Assume that 30 students registered for the course.

This means that [tex]n = 30[/tex]

Compute the probability that 2 or fewer will withdraw:

This is:

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

In which

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

[tex]P(X = 0) = C_{30,0}.(0.25)^{0}.(0.75)^{30} = 0.0002[/tex]

[tex]P(X = 1) = C_{30,1}.(0.25)^{1}.(0.75)^{29} = 0.0018[/tex]

[tex]P(X = 2) = C_{30,2}.(0.25)^{2}.(0.75)^{28} = 0.0086[/tex]

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0002 + 0.0018 + 0.0086 = 0.0106[/tex]

0.0106 = 1.06% probability that 2 or fewer will withdraw

Answer:

[tex](f + g)(4) = 191[/tex]

Step-by-step explanation:

Given

[tex]f(x) = 5x^2 - 5x + 15[/tex]

[tex]g(x) = 6x^2 + 7x - 8[/tex]

Required

[tex](f + g)(4)[/tex]

First, calculate [tex](f + g)(x)[/tex]

This is calculated as:

[tex](f + g)(x) = f(x) + g(x)[/tex]

So, we have:

[tex](f + g)(x) = 5x^2 - 5x + 15+6x^2 + 7x - 8[/tex]

Collect like terms

[tex](f + g)(x) = 5x^2 +6x^2 - 5x+ 7x + 15 - 8[/tex]

[tex](f + g)(x) = 11x^2 + 2x + 7[/tex]

Substitute 4 for x

[tex](f + g)(4) = 11*4^2 + 2*4 + 7[/tex]

[tex](f + g)(4) = 191[/tex]

(3x-2)(3x+ 2)

Multiply each term in one set of parentheses by the other terms:

3x x 3x =9x^2

3x x 2 = 6x

-2 x 3x = -6x

-2 x 2 = -4

Combine to get:

9x^2 + 6x - 6x -4

Combine like terms:

9x^2 - 4

The value of a would be 9.

[tex]\sf{\bold{\blue{\underline{\underline{Given}}}}}[/tex]

[tex]\sf{\bold{\red{\underline{\underline{To\:Find}}}}}[/tex]

[tex]\sf{\bold{\purple{\underline{\underline{Solution}}}}}[/tex]

At first we have to solve the value of (3x-2)(3x+2)

According to the question,

⠀⠀⠀

[tex]\sf{\bold{\green{\underline{\underline{Answer}}}}}[/tex]

Hence,

The value of a is 9

Answer:

is it going to be 10.5

Step-by-step explanation:

I do not have any explanation

Answer: 0 (zero)

Step-by-step explanation:

Start Learning & start growing! edge2023

*DROPS THE MIC*

Answer:

3/8

Step-by-step explanation:

the total number of possible results is 4×4=16.

out of these 16 only the results

1 2

1 3

1 4

2 2

2 3

3 2

are desired results. these are 6.

so the probability of a desired result is 6/16 = 3/8

Answer:

z = ±  0.772193214

Step-by-step explanation:

Hence, the critical value for a [tex]88[/tex]% confidence level is [tex]z=1.56[/tex].

What is the standard deviation?

Standard deviation is the degree of dispersion or the scatter of the data points relative to its mean, in descriptive statistics.

It tells how the values are spread across the data sample and it is the measure of the variation of the data points from the mean.

Here given that,

For a confidence level of  [tex]88[/tex]%, find the critical value for a normally distributed variable.

Let us assume that the standard normal distribution having a mean is [tex]0[/tex] and the standard deviation is [tex]1[/tex].

As the significance level is [tex]1[/tex] - confidence interval

Confidence interval is [tex]\frac{80}{100}=0.88[/tex]

i.e., [tex]1-0.88=0.12[/tex]

For the two sided confidence interval the confidence level is [tex]0.44[/tex].

Now, the standard normal probability table the critical value for the  [tex]88[/tex]% confidence level is [tex]1.56[/tex].

Hence, the critical value for a [tex]88[/tex]% confidence level is [tex]z=1.56[/tex].

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There are more compact cars (4*10 = 40) compared to trucks (2*10 = 20); however, the pictogram might make it appear that there are more trucks because the individual truck icon is larger compared to an individual compact car icon.

To anyone giving this image a quick glance, they may erroneously conclude that there are more trucks since their eye would notice the trucks first. Also, the person might think there are more trucks because bigger sizes tend to correspond to more proportion.

In real life, a truck is larger than a compact car, but the icons need to be the same size to have the figure not be misleading.

A very similar issue happens with the mid-size cars vs the compact cars as well. The three mid-size car icons span the same total width as the compact cars do, indicating that a reader might mistakenly conclude that there are the same number of mid-size cars compared to compact ones (when that's not true either).

Answer:

Quadratic formula

Step-by-step explanation:

The function is quadratic because it is a parabola. Exponential functions shoot either upwards or downwards rapidly, and it is clearly not linear due to it's curve. It also isn't piecewise because the function never stops or starts irregularly.

Answer:

2 days

Step-by-step explanation:

Lets say that in one day, one cow eats 1 block of grass. So, six cows in three days would eat 18 blocks of grass in total. So 18 blocks of grass is how much we have. that means nine cows would eat that much in 2 days.

Answer:

AnBnC ( common place for all)

HAVE A NİCE DAY

Answer:

D. 750(0.75)ˣ

Step-by-step explanation:

Let the new brightness be L'. Since our initial brightness L₀ reduces by 25 %, we have that L' = L₀ - 25% of L₀

L' = L₀ - 0.25L₀

L' = 0.75L₀

Adding the second screen, the new intensity is L" = L' - 25 % of L'  

L" = L' - 0.25 L'

L" = 0.75L'.

Since L' = 0.75L₀,

L" = 0.75L' = 0.75(0.75L₀) = 0.75²L₀

Adding the third screen, the new intensity is L"' = L'' - 25 % of L''  

L'" = L" - 0.25 L"

L"' = 0.75L".

Since L" = 0.75L' = 0.75²L₀

L"' = 0.75L" = 0.75(0.75²L₀) = 0.75³L₀

So, we see a pattern here.

The intensity after x screens is L = (0.75)ˣL₀

Since L₀ = 750 lumens,

L = 750(0.75)ˣ

Hello!

[tex]\large\boxed{P = 300m}[/tex]

Use the following formula for the perimeter:

P = 2l + 2w, where:

l = length

w = width

Therefore:

P = 2(90) + 2(60)

Simplify:

P = 180 + 120 = 300 m

Answer:

well how about you use common sense 100 yards long on each side 200 yards then add 5o yards since the the that is how wide it is then add another 50 and you get 300 yards then convert that to meters

Answer:

[tex]\begin{array}{ccc}{Class} & {Frequency} & {Relative\ Frequency} &{30-39} & {1} & {0.025} & {40-49} & {1} & {0.025} & {50 - 59} & {2} & {0.050} & {60 - 69} & {5} & {0.125} & {70 - 79} & {13} & {0.325} & {80 - 89} & {10} & {0.250} & {90 - 99} & {8} & {0.200} &{Total} & {40} & {1}\ \end{array}[/tex]

[tex]P(x < 70) = 0.225[/tex]

Step-by-step explanation:

Given

[tex]Lower = 30[/tex]

[tex]Width = 10[/tex]

Solving (a): The relative frequency table

First, we construct the frequency table using the given parameters.

[tex]\begin{array}{cc}{Class} & {Frequency} &{30-39} & {1} & {40-49} & {1} & {50 - 59} & {2} & {60 - 69} & {5} & {70 - 79} & {13} & {80 - 89} & {10} & {90 - 99} & {8} & {Total} & {40}\ \end{array}[/tex]

The relative frequency (RF) is calculated as:

[tex]RF = \frac{Frequency}{Total}[/tex]

Using the above formula to calculate the relative frequency, the relative frequency table is:

[tex]\begin{array}{ccc}{Class} & {Frequency} & {Relative\ Frequency} &{30-39} & {1} & {0.025} & {40-49} & {1} & {0.025} & {50 - 59} & {2} & {0.050} & {60 - 69} & {5} & {0.125} & {70 - 79} & {13} & {0.325} & {80 - 89} & {10} & {0.250} & {90 - 99} & {8} & {0.200} &{Total} & {40} & {1}\ \end{array}[/tex]

Solving (b): [tex]P(x < 70)[/tex]

To do this, we add up the relative frequencies of classes less than 70.

i.e.

[tex]P(x < 70) = [30 - 39] + [40 - 49] + [50 - 59] + [60 - 69][/tex]

So, we have:

[tex]P(x < 70) = 0.025 + 0.025 + 0.050 + 0.125[/tex]

[tex]P(x < 70) = 0.225[/tex]

Answer:

prove that The given system of equations is redundant is attached below

Step-by-step explanation:

System of equations

x1 +x2 −3x3 = 7

−2x1 +x2 +5x3 = 2

 3x2 −x3 = 16

To prove that the system is redundant we will apply the Gaussian elimination ( pivot operation )

attached below is the solution

Answer:

D) y = –5x + 4x^2 + 2

Step-by-step explanation:

You can tell by the first number being positive or negative. To check use Desmo graphing calculator and enter your equation for next time.

9514 1404 393

Answer:

  66 m

Step-by-step explanation:

The perimeter is the sum of the measures of all of the sides. There are two side measures that are missing from the diagram.

The missing horizontal measure is ...

  17 m - 8 m = 9 m

The missing vertical measure is ...

  16m -7 m = 9 m.

If you add these to the sum you already calculated, you will get the correct answer:

  48 m + 9 m + 9 m = 66 m . . . perimeter of the figure

_____

If you're paying attention, you see that the sum of the measures of the two shorter horizontal segments is the same as the measure of the longer horizontal segment. Likewise, the sum of the measurements of the two shorter vertical segments is the same as that of the longer vertical segment.

In other words, the perimeter of this (and any) L-shaped figure is the same as the perimeter of a rectangle having the same horizontal and vertical dimensions as the long sides of the figure.

  P = 2(17 m +16 m) = 2(33 m) = 66 m

Answer:

The answer is "840".

Step-by-step explanation:

Following are the number of ways in which selecting a team A  by 9 children:

[tex]= ^{9_{C_{3}}\\\\\\[/tex]

[tex]=\frac{9!}{3! \times 6!} \\\\=\frac{9\times 8\times 7\times 6!}{3 \times 2\times 1\times 6!}\\\\=\frac{9\times 8\times 7}{3 \times 2\times 1}\\\\=\frac{3\times 4\times 7}{1}\\\\=\frac{84}{1}\\\\=84[/tex]

Following are the number of ways in which selecting a team B  by remaining 6 children:

 [tex]= ^{6}_{C_{3}}[/tex]

[tex]= \frac{6!}{(3! \times 3!)}\\\\= \frac{6!}{(3\times 2\times 1 \times 3!)}\\\\= \frac{6\times 5 \times 4 \times 3!}{(3\times 2\times 1 \times 3!)}\\\\= \frac{ 5 \times 4 \times 3!}{3!}\\\\= 5 \times 4 \\\\=20[/tex]

Following are the number of ways in which selecting a team C  by remaining 3 children:

[tex]= ^{3}_{C_{3}}\\\\=\frac{3!}{3!}\\\\= 1[/tex]

Following are the number of ways in which making 3 teams by 9 children:

[tex]= \frac{(84 \times 20 \times 1)}{3!}\\\\= \frac{(84 \times 20 )}{6}\\\\= 14 \times 20\\\\= 280\\\\[/tex]

(Note: we've split by 3! Because it also is necessary to implement three teams between themselves)

Now 3 leagues have to be played. One is going to be run by each team.

That is the way it is

Different possible divisions

[tex]= 280 \times 3!\\\\= 280 \times (3 \times 2 \times 1)\\\\= 840[/tex]