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3 years ago

Why might you need measure a perimeter with a measuring tape instead of a ruler

Mathematics

2 answers:

Anuta_ua [19.1K]3 years ago

8 0

Because a ruler is 4 times smaller minimum of mesuring tape

hammer [34]3 years ago

3 0

The perimeter may not be small enough to measure with just a ruler. For example, if a side of a rectangle was 30 inches, how would you use a 12 inch ruler to accurately measure that side? Also, if you were measuring a shape like a circle or oval, a tape measure is flexible.

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There are 15 blueberry muffins in a large basket. This represents 5/9 of all the muffins that are in the basket. The equation 5/

noname [10]

The answer is 27 muffins.

5/9m=15

9/5*5/9 which cancels each other out leaving just m.

Then you multiply 15 by 9/5 which gives you 135/5 which reduces to 27.

7 0

3 years ago

Read 2 more answers

A rectangle is transformed according to the rule R0, 90º. The image of the rectangle has vertices located at R'(–4, 4), S'(–4, 1

zalisa [80]

We are said that a rectangle has been transformed into the one indicated in Figure 1 according to this rule:

$R_0, \ 90^{\circ}$

We know that the center of rotation is origin and two rules are applied to rotate a point 90 degrees, namely:

1. Clockwise

In this case, the rule to transform a point is:

$(x,y) \rightarrow (y,-x)$

This rule was already applied to form the image, so all we need to do is to reverse the answer using this formula, therefore:

$For \ Q(-3,4): \\ \\ (y,-x)=(-3,4) \\ \\ \therefore y=-3 \ and \ -x=4 \therefore x=-4 \\ \\ Thus, \ the \ point \ is: \\ \\ \boxed{Q(-4,-3)}$

2. Counterclockwise

Applying the same previous concept but with the new rules for this case:

$(x,y) \rightarrow (-y,x)$

By reversing the answer, we have:

$For \ Q(-3,4): \\ \\ (-y,x)=(-3,4) \\ \\ \therefore -y=-3 \therefore y=3 \ and \ x=4 \\ \\ Thus, \ the \ point \ is: \\ \\ \boxed{Q(4,3)}$

3 0

2 years ago

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A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round

masha68 [24]

Answer:

On the way up, 1.8 seconds and on the way down 4.1 seconds.

Step-by-step explanation:

Use the motion under gravity equation s = ut + 0.5gt^2 where u = initial velocity, s = the height, t = time and g = acceleration due to gravity ( 32 fts-2).

So 120 = 95t + 0.5(-32)t^2

-16t^2 + 95t - 120 = 0

Solving this using the quadratic formula:

t = [-95 +/- √(95^2 - 4 * -16*-120)] / -32

t = 1.823, 4.1148.

The first value refers to the upward journey and the second to the downward journey.

4 0

3 years ago

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Four times Theo’s number added 20 is 10. what is theo’s number ?

stellarik [79]

Theos number is -2/5

4x + 20 = 10
-20 -20
4x = -10
x = -4/10
x = -2/5

4 0

3 years ago

Read 2 more answers

Suppose that wait times for customers at a grocery store cashier line are uniformly distributed between one minute and twelve mi

olchik [2.2K]

a) Mean: 6.5 min, variance: 10.1 min

b) 0.54 (54%)

c) 0.73 (73%)

d) 0.34

Step-by-step explanation:

Here we can call X the variable indicating the waiting time for the customers:

X = waiting time

We are told that the waiting time is distributed uniformly between 1 and 12; this means that

$1\leq X \leq 12$

And the probability is equal for each value of X, so:

$p(X=1)=p(X=2)=....=p(X=12)$

The mean of a uniform distribution is given by:

$E[X]=\frac{b+a}{2}$

where a and b are the minimum and maximum values of the variable X. In this case,

a = 1

b = 12

So the mean value of X is

$E[X]=\frac{12+1}{2}=6.5$ (minutes)

The variance of a uniform distribution is given by:

$Var[X]=\frac{1}{12}(b-a)^2$

And substituting the values of this problem,

$Var[X]=\frac{1}{12}(12-1)^2=10.1$ (minutes)

Since the distribution is uniform between 1 and 12, we can write the probability density function as

$f(x)=\frac{1}{b-a}$

The cumulative function gives the probability that the values of X is less than a certain value t:

$p(X$ (1)

In this case, we want to find the probability that the waiting time is less than 7 minutes, so

t = 7

We also have:

a = 1

b = 12

Therefore, calculating (1) and substituting, we find:

$p(X$

The probability that a customer waits between four and twenty minutes can be rewritten as

$p(4$

This can be written as:

$p(4$ (1)

However, the probabilty of X>4 can be written as

$p(X>4)=1-p(X$

Also, we notice that

$p(X$ because the maximum value of X is 12; therefore, we can rewrite (1) as

$p(4$

We can calculate $p(X$ by using the same method as in part b:

$p(X$

So, we find

$p(4$

In this part, we know that a customer waits for

X = k

minutes in line, and he receives a coupon worth

$0.2k^{\frac{1}{4}}$ dollars.

Here we want to find the mean of the coupon value.

Here therefore we have a new variables defined as

$Y=0.2X^{\frac{1}{4}}$

Given a variable with standard (between 0 and 1) uniform distribution X, the variable

$Y=X^n$

follows a beta distribution, with parameters $(\frac{1}{n},1)$ , and whose mean value is given by

$E[Y]=\frac{1/n}{1+\frac{1}{n}}$

In this case,

$n=\frac{1}{4}$

So the mean value of $X^{1/4}$ is

$E[X^{1/4}]=\frac{1/(1/4)}{1+\frac{1}{1/4}}=\frac{4}{1+4}=\frac{4}{5}=0.8$

However, our variable is distribution is non-standard, because its values are between 1 and 12, so the range is

$Min = 1^{1/4}=1\\Max =12^{1/4}=1.86$

So, the actual mean value of $X^{1/4}$ is

$E[X^{1/4}]=0.8\cdot (1.86-1)+1=1.69$

However, in the definition of Y we also have a factor 0.2; therefore, the mean value of Y is

$E[Y]=0.2E[X^{1/4}]=0.2\cdot 1.69 =0.34$

5 0

3 years ago