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2 years ago

11

A fitness club holds an open house for new members. The club expects 6,500 people to sign up for gym membership. If there is a p

ercent error of 4%, what is the range of the number of people who will sign up for gym membership?
Enter the correct answers in the boxes.

Least number of people:

Greatest number of people:

2 answers:

Nata [24]2 years ago

5 0

Firstly, 6500 divided by 100 then x by 4= 260, 260 + 6500= 6760.
The greatest amount is 6760
Then you do 6500-260= 6240
The least amount is 6240

Sunny_sXe [5.5K]2 years ago

3 0

The answer would be 6240 have a good day!

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. For each of these intervals, list all its elements or explain why it is empty. a) [a, a] b) [a, a) c) (a, a] d) (a, a) e) (a,

Eva8 [605]

Answer:

Elements are of the form

(i) $[a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}$

(ii) $[a,b)=\{[x,y) :a\leq x$

(iii) $(a,a]=\{(x,y] :a$

(iv) $(a,a)=\{(x,y): a$

(v) (a,b) where a>b= $\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}$

(vi) [a,b] where a>b= $\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}$

Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi) [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

$\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a] represents singleton sets, and singleton sets are empty so is [a,a].

(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

$[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........$

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

$( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

(iv) (a,a)

Means given set excluding a by left as well as right.

Since there is a singleton element a of real numbers, this set is empty.

Its elements are of the form.

$( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Because there is no increment so if $a\in \mathbb R$ then the set (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).

(v) (a,b) where a>b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$(a,b)=\{(5,0),(5,1),(5,2).....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

(vi) [a,b] where $a\leq b$ .

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$[a,b]=\{[5,0],[5,1],[5,2].....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

8 0

2 years ago

Aaron, Blaine, and Cruz are solving the equation 4 7 (7 − n) = −1. Aaron started his solution by multiplying both sides of the e

hjlf

Answer:

D. All three chose a valid first step toward solving the equation.

Step-by-step explanation:

Aaron, Blaine, and Cruz are solving the equation 4/7 (7 − n) = −1. Aaron started his solution by multiplying both sides of the equation by 7/4 . Blaine started by using the distributive property to multiply 4/7 by both 7 and −n. Cruz started by dividing both sides of the equation by 4/7 .

Which of the following is true?

A. Blaine and Cruz made an error in picking their first steps.

B. Cruz made an error in picking his first step.

C. All three made an error because the right side equals -1.

D. All three chose a valid first step toward solving the equation.

Given:

4/7(7 - n) = -1

Aaron:

4/7(7 - n) = -1

Multiple both sides by 7/4

4/7(7 - n) * 7/4 = -1 * 7/4

7 - n = -7/4

- n = -7/4 - 7

- n = (-7-28)/4

- n = -35/4

n = 35/4

Blaine:

4/7(7 - n) = -1

4/7(7 - n) × 7 = -1 × 7

4(7 - n) = -7

28 - 4n = -7

-4n = -7 - 28

- 4n = - 35

n = -35/-4

n = 35/4

Cruz:

4/7(7 - n) = -1

Divide both sides by 4/7

4/7(7 - n) ÷ 4/7 = -1 ÷ 4/7

4/7(7 - n) × 7/4 = -1 × 7/4

7 - n = -7/4

- n = (-7-28)/4

- n = -35/4

n = 35/4

D. All three chose a valid first step toward solving the equation.

7 0

2 years ago

exis [7]

Answer:

Option C. x = 5

Step-by-step explanation:

To solve this question we should always keep these two points in our mind.

1). In-center of a triangle is formed by the intersection of the angle bisectors of all interior angles of the triangle.

2). In-center of a triangle is equidistant from all three sides of the triangle.

From the figure attached,

In ΔSRT,

∠ASZ ≅ ∠ZSB [ SZ is an angle bisector of angle ASB]

5x - 9 = 16 [ given in the question]

5x = 16 + 9

5x = 25

x = $\frac{25}{5}$

x = 5

Therefore, Option C. is the answer.

5 0

2 years ago

Read 2 more answers

The value of a house depreciates by 6% per year. Work out the current value of a house bought 3 years ago for £250000.

ss7ja [257]

$\bf \qquad \textit{Amount for Exponential Decay}
\\\\
A=P(1 - r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &250000\\
r=rate\to 6\%\to \frac{6}{100}\to &0.06\\
t=\textit{elapsed time}\to &3\\
\end{cases}
\\\\\\
A=250000(1-0.06)^3\implies A=250000(0.94)^3\implies A=207646$

5 0

2 years ago

Use the definition to find an expression for the area under the curve y = x3 from 0 to 1 as a limit. lim n→∞ n i = 1 R (b) The f

Harlamova29_29 [7]

The summand (R?) is missing, but we can always come up with another one.

Divide the interval [0, 1] into $n$ subintervals of equal length $\dfrac{1-0}n=\dfrac1n$ :

$[0,1]=\left[0,\dfrac1n\right]\cup\left[\dfrac1n,\dfrac2n\right]\cup\cdots\cup\left[1-\dfrac1n,1\right]$

Let's consider a left-endpoint sum, so that we take values of $f(\ell_i)={\ell_i}^3$ where $\ell_i$ is given by the sequence

$\ell_i=\dfrac{i-1}n$

with $1\le i\le n$ . Then the definite integral is equal to the Riemann sum

$\displaystyle\int_0^1x^3\,\mathrm dx=\lim_{n\to\infty}\sum_{i=1}^n\left(\frac{i-1}n\right)^3\frac{1-0}n$

$=\displaystyle\lim_{n\to\infty}\frac1{n^4}\sum_{i=1}^n(i-1)^3$

$=\displaystyle\lim_{n\to\infty}\frac1{n^4}\sum_{i=0}^{n-1}i^3$

$=\displaystyle\lim_{n\to\infty}\frac{n^2(n-1)^2}{4n^4}=\boxed{\frac14}$

8 0

2 years ago