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2 years ago

8

Candy draws a square design with a side length of x inches for the window at the pet shop. She takes the design to the printer a

nd asks for a sign that has an area of 16x2 – 40x + 25 square inches.
What is the side length, in inches, of the pet shop sign?

4x + 5
4x – 5
8x + 5
8x – 5

2 answers:

Inga [223]2 years ago

8 0

Answer:

This is late, but for anyone else that needs it, it's B. 4x-5

Harrizon [31]2 years ago

7 0

Answer:

B. $(4x-5)$

Step-by-step explanation:

We are given that,

Area of the square design = $16x^2 -40x + 25$ inch²

Factorizing the polynomial, we see that,

$16x^2 -40x + 25=0$

implies $(4x-5)^2=0$

So, the factors of the polynomial are $(4x-5)$ and $(4x-5)$

That is, $16x^2 -40x + 25=0$ = $(4x-5)^2=0$ .

Thus, we have,

Area of the square design = $16x^2 -40x + 25$ = $(4x-5)^2$

As, Area of a square = $Side^2$

Hence, on comparing the sides of the square design are $(4x-5)$ .

So, option B is correct.

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Credit cards place a three-digit security code on the back of cards. What is the probability that a code starts with the number

ohaa [14]

Answer: 1/10

Step-by-step explanation: because there are 10 numbers (1,2,3,4,5,6,7,8,9, and 0)

7 0

2 years ago

Which of the following are dimensionally consistent? (Choose all that apply.)(a) a=v / t+xv2 / 2(b) x=3vt(c) xa2=x2v / t4(d) x=v

Bumek [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

is dimensionally consistent

B

is not dimensionally consistent

C

is dimensionally consistent

D

is not dimensionally consistent

E

is not dimensionally consistent

F

is dimensionally consistent

G

is dimensionally consistent

H

is not dimensionally consistent

Step-by-step explanation:

From the question we are told that

The equation are

$A) \ \ a^3 = \frac{x^2 v}{t^5}$

$B) \ \ x = t$

$C \ \ \ v = \frac{x^2}{at^3}$

$D \ \ \ xa^2 = \frac{x^2v}{t^4}$

$E \ \ \ x = vt+ \frac{vt^2}{2}$

$F \ \ \ x = 3vt$

$G \ \ \ v = 5at$

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$

Generally in dimension

x - length is represented as L

t - time is represented as T

m = mass is represented as M

Considering A

$a^3 = (\frac{L}{T^2} )^3 = L^3\cdot T^{-6}$

and $\frac{x^2v}{t^5 } = \frac{L^2 L T^{-1}}{T^5} = L^3 \cdot T^{-6}$

Hence

$a^3 = \frac{x^2 v}{t^5}$ is dimensionally consistent

Considering B

$x = L$

and

$t = T$

Hence

$x = t$ is not dimensionally consistent

Considering C

$v = LT^{-1}$

and

$\frac{x^2 }{at^3} = \frac{L^2}{LT^{-2} T^{3}} = LT^{-1}$

Hence

$v = \frac{x^2}{at^3}$ is dimensionally consistent

Considering D

$xa^2 = L(LT^{-2})^2 = L^3T^{-4}$

and

$\frac{x^2v}{t^4} = \frac{L^2(LT^{-1})}{ T^5} = L^3 T^{-5}$

Hence

$xa^2 = \frac{x^2v}{t^4}$ is not dimensionally consistent

Considering E

$x = L$

;

$vt = LT^{-1} T = L$

and

$\frac{vt^2}{2} = LT^{-1}T^{2} = LT$

Hence

$E \ \ \ x = vt+ \frac{vt^2}{2}$ is not dimensionally consistent

Considering F

$x = L$

and

$3vt = LT^{-1}T = L$ Note in dimensional analysis numbers are

not considered

Hence

$F \ \ \ x = 3vt$ is dimensionally consistent

Considering G

$v = LT^{-1}$

and

$at = LT^{-2}T = LT^{-1}$

Hence

$G \ \ \ v = 5at$ is dimensionally consistent

Considering H

$a = LT^{-2}$

,

$\frac{v}{t} = \frac{LT^{-1}}{T} = LT^{-2}$

and

$\frac{xv^2}{2} = L(LT^{-1})^2 = L^3T^{-2}$

Hence

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$ is not dimensionally consistent

8 0

2 years ago

Scores on Ms. Bond's test have a mean of 70 and a standard deviation of 11. David has a score of 52 on Ms. Bond's test. Scores o

otez555 [7]

Answer:

Due to the higher z-score, David has the higher standardized score

Step-by-step explanation:

Z-score:

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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.

Which student has the higher standardized score

Whoever had the higher z-score.

David:

Scores on Ms. Bond's test have a mean of 70 and a standard deviation of 11. David has a score of 52 on Ms. Bond's test. So $X = 52, \mu = 70, \sigma = 11$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{52 - 70}{11}$

$Z = -1.64$

Steven:

Scores on Ms. Nash's test have a mean of 64 and a standard deviation of 6. Steven has a score of 52 on Ms. So $X = 52, \mu = 64, \sigma = 6$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{52 - 64}{6}$

$Z = -2$

Due to the higher z-score, David has the higher standardized score

3 0

2 years ago

A probability calculator is required on this problem; answer to six decimal places. Suppose we will spin the wheel pictured 400

KiRa [710]

Answer:

$P(90< X< 110)= P(\frac{90-80}{8}$

And we can find this probability with this difference:

$P(90< X< 110)=P(z$

And we can find the real value with the following excel code using the binomial distribution:

"=BINOM.DIST(110,400,0.2,TRUE)-BINOM.DIST(89,400,0.2,TRUE)"

And we got 0.118 a very close value from the value obtained using the normal approximation

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=400, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

We need to check the conditions in order to use the normal approximation.

$np=400*0.2=80 \geq 10$

$n(1-p)=400*(1-0.2)=320 \geq 10$

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

$E(X)=np=400*0.2=80$

$\sigma=\sqrt{np(1-p)}=\sqrt{400*0.2(1-0.2)}=8$

So then we can approximate the random variable as $X \sim N(\mu = 80, \sigma = 8)$

And we want this probability:

$P(90< X< 110)$

We can use the z score formula given by:

$z = \frac{x -\mu}{\sigma}$

And replacing we got:

$P(90< X< 110)= P(\frac{90-80}{8}$

And we can find this probability with this difference:

$P(90< X< 110)=P(z$

And we can find the real value with the following excel code using the binomial distribution:

"=BINOM.DIST(110,400,0.2,TRUE)-BINOM.DIST(89,400,0.2,TRUE)"

And we got 0.118 a very close value from the value obtained using the normal approximation

8 0

2 years ago

The graphs below have the same shape. What is the equation for the red graph?

Serhud [2]

Your answer would be B because even though it's the same shape the red shape is 2 times bigger.......your exponent 4 stays 4

8 0

2 years ago

Read 2 more answers