All categories

3 years ago

Find the area of the figure. 7 cm 4 cm 4 cm 1 10 cm 10 cm 8 cm 19 cm

Mathematics

1 answer:

Paladinen [302]3 years ago

3 0

Answer:

132cm^2

Step-by-step explanation:

first break the shape up into component shapes. the shape is made of two rectangles and two triangles. the area for a rectangle is A=BH where b is base and h is height. the area of a right triangle is BH/2 where b is the base and h is the height. you are given the base and height of the top rectangle 4x7. To find the area of the triangles and the second rectangle you have to do some algebra. the total base length is 19cm and the top rectangle has a length of 7cm so 19-7 leaves you with 12cm to spare and there is two symmetric triangles on each side so 12/2 is 6. now we know all the measurements and can solve.

You might be interested in

#1: Simplify the expression below. Type your answer as an integer.<br> 7 + 1 - 18 : 6

myrzilka [38]

Answer:

Step-by-step explanation:

<u>Steps of calculation:</u>

7 + 1 - 18 : 6 =
7 + 1 - 3 =
8 - 3 =
5

Answer is 5

8 0

3 years ago

Suppose you pick two cards from a deck of 52 playing cards. What is the probability that they are both queens?

photoshop1234 [79]

Answer:

0.45% probability that they are both queens.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes

The combinations formula is important in this problem:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

Desired outcomes

You want 2 queens. Four cards are queens. I am going to call then A,B,C,D. A and B is the same outcome as B and A. That is, the order is not important, so this is why we use the combinations formula.

The number of desired outcomes is a combinations of 2 cards from a set of 4(queens). So

$D = C_{4,2} = \frac{4!}{2!(4-2)!} = 6$

Total outcomes

Combinations of 2 from a set of 52(number of playing cards). So

$T = C_{52,2} = \frac{52!}{2!(52-2)!} = 1326$

What is the probability that they are both queens?

$P = \frac{D}{T} = \frac{6}{1326} = 0.0045$

0.45% probability that they are both queens.

4 0

3 years ago

What is the answer Simplify. 1/4(1-2/3 squared+1/3

NemiM [27]

The Answer is 2/3

Steps: 1/4(1-(2/3^2+1/3)

3 0

3 years ago

Line JK passes through points J(–4, –5) and K(–6, 3). If the equation of the line is written in slope-intercept form, y = mx + b

4vir4ik [10]

The value of "b" is the y-intercept.

In order to figure out slope-intercept form you need 1 coordinate and the slope.
1) Find the slope, using the 2-point slope formula: "m= y2-y1 / x2-x1".
ex. m= -5 - 3 / -4 - -6 (simplify)---> m= -4

2) Fill in the blanks for point-slope formula: "y - y1 = m (x - x1)"
(choose one coordinate, it doesn't matter which one)
ex. y - -5 = -4 (x - -4)

3) Then use basic algebra to simplify.

7 0

3 years ago

The local swim team is considering offering a new semi-private class aimed at entry-level swimmers, but needs a minimum number o

SCORPION-xisa [38]

Answer:

The significance level is $\alpha=0.01$ and since we are conducting a right tailed test we need to find a critical value who accumulate 0.01 of the area in the right of the normal standard distribution and we got:

$z_{\alpha/2}= 2.326$

So we reject the null hypothesis is $z>2.326$

Step-by-step explanation:

For this case we define the random variable X as the number of entry-level swimmers and we are interested about the true population mean for this variable . On specific we want to test this:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

And the statistic is given by:

$z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

$z_{\alpha/2}= 2.326$

So we reject the null hypothesis is $z>2.326$

7 0

3 years ago