The length of a rectangular picture is 17 inches and the perimeter is 60 inches. find the width

Saul decides to use the IQR to measure the spread of the data. Saul calculates the IQR of the data set to be 27

german

Answer:

n = 78

Step-by-step explanation:

IQR is the interquartile range of the data set.

So Saul had already arranged the data in order, that is the first step when calculating the interquartile range.

25, 30, 50, 50, 50, 50, 56, n., 250

Next, is to find the median so we could easily separate the data into quartiles.

The median of the data 25, 30, 50, 50, 50, 50, 56, n., 250 = 50

Then, arrange the data into the first quartile and the third quartile

first quartile = 25, 30, 50, 50

third quartile = 50, 56, n., 250

first quartile median = 30 + 50 = 80/2 = 40

third quartile = 56 + n /2

The interquartile range formula is the first quartile subtracted from the third quartile:

IQR = Q3 – Q1.

Since 27 is the interquartile range, n=

27 = (56 + n /2) - 40

Use 2 to multiply through to eliminate the denominator

54 = (56 + n) - 80

n = 78

5 0

3 years ago

State if the two triangles are congruent. If they are, state how you know.

aksik [14]

Answer:

Yes they are.

They are congruent because they have two equal angles and one equal side

7 0

2 years ago

One more dilation problem <br> Write the rule to describe each transformation??

Leto [7]

Answer:

It is stretch.

I hope it helps a little bit.

8 0

2 years ago

Read 2 more answers

Use the following steps to prove that log b(xy)- log bx+ log by.

borishaifa [10]

Answer with Step-by-step explanation:

a. $x=b^p$

$y=b^q$

Taking both sides log

$log x=plog b$

Using identity: $logx^y=ylogx$

$p=\frac{logx}{log b}=log_b x$

Using identity: $log_x y=\frac{log y}{log x}$

$log y=qlog b$

$q=\frac{log y}{log b}=log_b y$

b. $xy=b^pb^q$

We know that

$x^a\cdot x^b=x^{a+b}$

Using identity

$xy=b^{p+q}$

c. $log_b(xy)=log_b(b^{p+q})$

$log_b(xy)=(p+q)log_b b$

Substitute the values then we get

$log_b(xy)=(log_b x+log_b y)$

By using $log_b b=1$

Hence, $log_b(xy)=log_b x+log_b y$

3 0

3 years ago

Find a point on the ellipsoid x2+y2+4z2=36x2+y2+4z2=36 where the tangent plane is perpendicular to the line with parametric equa

Anvisha [2.4K]

Answer:

A point on the ellipsoid is (-4,2,2) or (4,-2,-2)

Step-by-step explanation:

Given equation of ellipsoid f(x,y,z) : $x^2+y^2+4z^2=36$

Parametric equations:

x=-4t-1

y=2t+1

z=8t+3

Finding the gradient of function

$\nabla f(x,y,z)=\\\nabla f(x,y,z)=$

So, The directions vectors=(-4,2,8)

Now the line is perpendicular to plane when direction vector is parallel to the normal vector of line

$\nablaf(x,y,z)=(2x,2y,8z)=\lambda(-4,2,8)$

So, $2x=-4\lambda$

$\Rightarrow x=-2\lambda$

$2y=2\lambda\\\Rightarrow y=\lambda\\8z=8\lambda\\\Rightarrow z=\lambda$

Substitute the value of x , y and z in the ellipsoid equation

$(2\lambda)^2+(\lambda)^2+4(\lambda)^2=36\\9(\lambda)^2=36\\\lambda^2=4\\\lambda=\pm 2$

With $\lambda = 2$

x=-2(2)=-4

y=2

z=2

With $\lambda =- 2$

x=-2(-2)=4

y=-2

z=-2

Hence a point on the ellipsoid is (-4,2,2) or (4,-2,-2)

7 0

3 years ago