All categories

3 years ago

PLS HELP ME I NEED HELP

Mathematics

1 answer:

Brut [27]3 years ago

7 0

Answer:

≈ 7.71 ft

Step-by-step explanation:

The perimeter is half the circumference (C) of the circle plus the diameter.

$\frac{1}{2}$ C = $\frac{1}{2}$ πd ( d is the diameter )

= 0.5 × π × 3 = 1.5π

Thus

perimeter = 1.5π + 3 ≈ 7.71 ft ( to the nearest hundredth )

You might be interested in

Suppose ARB Bank is reviewing its service charges and interest payment policies on current accounts. Suppose further that ARB ha

Advocard [28]

Answer:

5.94% of customers carries a balance of GH¢100 or lower.

82.64% of customers carries a balance of GH¢500 or lower.

0% of current account customers carries average daily balances exactly equal to GH¢500.

76.7% of customers maintains account balance between GH¢100 and GH¢500

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

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.

In this question, we have that:

$\mu = 350, \sigma = 160$

What percentage of customers carries a balance of GH¢100 or lower?

This is the pvalue of Z when X = 100. So

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

$Z = \frac{100 - 350}{160}$

$Z = -1.56$

$Z = -1.56$ has a pvalue of 0.0594

5.94% of customers carries a balance of GH¢100 or lower.

What percentage of customers carries a balance of GH¢500 or lower?

This is the pvalue of Z when X = 500.

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

$Z = \frac{500 - 350}{160}$

$Z = 0.94$

$Z = 0.94$ has a pvalue of 0.8264

82.64% of customers carries a balance of GH¢500 or lower.

What percentage of current account customers carries average daily balances exactly equal to GH¢500?

In the normal distribution, the probability of finding a value exactly equal to X is 0. So

0% of current account customers carries average daily balances exactly equal to GH¢500.

What percentage of customers maintains account balance between GH¢100 and GH¢500?

This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 100.

From b), when X = 500, Z = 0.94 has a pvalue of 0.8264

From a), when X = 100, Z = -1.56 has a pvalue of 0.0594

0.8264 - 0.0594 = 0.767

76.7% of customers maintains account balance between GH¢100 and GH¢500

5 0

3 years ago

Imelda, Susan, and Clara are driving go-carts around a track. Imelda takes 14 minutes, Susan takes 18 minutes, and Clara takes 1

RoseWind [281]

Answer:

630 minutes.

Step-by-step explanation:

That would be the lowest common multiple of 10, 14 and 18

10 = 2 x 5

14= 2 x 7

18 = 2 x 3 x 3

The LCM = 2 x 3 x 3 x 5 x 7

= 630.

4 0

3 years ago

What is 0.057×10 equal

lbvjy [14]

0.057x 10 is equal to .57. All you have to do is move the decimal to the right one time because there is one zero in 10.

8 0

3 years ago

Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

Svetlanka [38]

Answer:

(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.

5 0

3 years ago

Scott takes a walk around the block. How far did he walk?

goldfiish [28.3K]

I’m pretty sure it’s 240 because you multiply then add

7 0

3 years ago