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2 years ago

Help ill give brainiest pls

Mathematics

2 answers:

Vanyuwa [196]2 years ago

5 0

Answer:

Step-by-step explanation:

Hops this helps

mihalych1998 [28]2 years ago

3 0

The answer to your question is D! i just took the quiz! D) Yes because aamjgle 6 and andgle 7 are congruent

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Rectangle with area of 10 and length of 2 1/2. Whats the width?

Stels [109]

Hello!

We have been given enough information to solve this question!
We know that the length of the rectangle is 2.5, and we also know how to find the area:

A = l × w

But we don't need the area, we need to find the width of the rectangle! You see, we had to multiply the length by the width to find the area.

Now, to find the width, we must flip the problem around and use the opposite of multiplication, which is division.

W = a ÷ l

W = 10 ÷ 2.5

W = 4

Therefore, the width of the rectangle is 4.

3 0

3 years ago

Read 2 more answers

imagine you live only one mile from work and you decide to walk.if you walk four milesb per hour how long will it take you to wa

olga nikolaevna [1]

1 mile * (1 hr / 4 mile) = 1/4 hr

1/4 hr = 15 min

5 0

4 years ago

Charlotte measures the distance from the center of a circular garden to its edge and finds it is 43 meters. How long is the path

yanalaym [24]

Answer: 270.29 meters

Step-by-step explanation:

This is a circle which means that the distance that Charlotte measured from the center of a circular garden to its edge is the radius.

The length of the path which goes around the garden would therefore be the circumference.

Circumference = π * diameter

Diameter = radius * 2

= 43 * 2

= 86 meters

Circumference = 22/7 * 86

= 270.29 meters

3 0

3 years ago

Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a

spin [16.1K]

Answer:

$Mean = 68.9$

$s^2 =18.1$ --- Variance

Step-by-step explanation:

Given

$\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

$x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5$

For 59 - 66

$x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5$

For 67 - 74

$x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5$

For 75 - 82

$x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5$

For 83 - 90

$x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5$

So, the table becomes:

$\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

The mean is then calculated as:

$Mean = \frac{\sum fx}{\sum f}$

$Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}$

$Mean = \frac{2547.5}{37}$

$Mean = 68.9$ -- approximated

Solving (b): The sample variance:

This is calculated as:

$s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}$

So, we have:

$s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}$

$s^2 =\frac{652.8}{36}$

$s^2 =18.1$ -- approximated

5 0

3 years ago

A jar containing only dimes and quarters contains a total of 54 coins. The value of all the coins in the jar is $9.15. Solve by

Paladinen [302]

Answer: the jar contains 29 dimes and 25 quarters.

Step-by-step explanation:

The worth of a dime is 10 cents. Converting to dollars, it becomes

10/100 = $0.1

The worth of a quarters is 25 cents. Converting to dollars, it becomes

25/100 = $0.25

Let x represent the number of dimes contained in the jar.

Let y represent the number of quarters contained in the jar.

A jar containing only dimes and quarters contains a total of 54 coins.. This means that

x + y = 54

The value of all the coins in the jar is $9.15. This means that

0.1x + 0.25y = 9.15 - - - - - - - - - - - 1

Substituting x = 54 - y into equation 1, it becomes

0.1(54 - y) + 0.25y = 9.15

5.4 - 0.1y + 0.25y = 9.15

- 0.1y + 0.25y = 9.15 - 5.4

0.15y = 3.75

y = 3.75/0.15 = 25

x = 54 - y = 54 - 25

x = 29

7 0

3 years ago