All categories

3 years ago

Find the area of the rhombus. Find the area. The figure is not drawn to scale.

Mathematics

2 answers:

Dmitry_Shevchenko [17]3 years ago

7 0

Answer:

I don't know if this is right but I think it is 32m

HACTEHA [7]3 years ago

6 0

Answer:

Hello! answer:

question 30: 128 m2

question 31: 1188 in2

Step-by-step explanation:

Hello just in advance I dont know if you need unuts or not but i believe those are correct if not Im sorry but the answers are 100% correct.

For the first one I found the area of a singular triangle then just multiplued by 4 cause all of those triangles are the same so... triangle area is... base × height ÷ 2 or base × height × 1/2

8 × 8 = 64 64 ÷ 2 = 32 32 × 4 = 128 therefore the area is 128 for the second I just did base × height therefore 36 × 33 = 1188 therefore thats the area hope that helps!

You might be interested in

ANSWER AND STEPS PLEASE. HELP

Ilia_Sergeevich [38]

Answer:

A. 75.36 in^3

Step-by-step explanation:

r =3 in

h=8 in

v=1/3 π r^2h

= 1/3×3.14×3×8

=75.36 in^2

hope it helps...

have a great day!!

8 0

3 years ago

The Target heart rate for fit person of age a exercising at p percent of his or her heart rate is determined by the expression 1

natita [175]

Answer: 9p - 1.5a

Step-by-step explanation:

Your maximum heart rate is about 220 minus your age. But it varies base on age category.

Target heart rate during moderate intensity activities is about 50-70% of maximum heart rate.

According to the question;

Target heart rate for

women = 1/2p(418-4a)

= 209p - 2a ....... (1)

men = 1/2p(400-a)

= 200p - 0.5a .......(2)

To find the difference between the target heart rates for a fit woman and a fit man, take equation 1 from equation 2

(209p - 2a) - (200p - 0.5a)

209p - 2a - 200p + 0.5a

9p -1.5a

7 0

4 years ago

Write an algebraic expression for the verbal expression. the product of 13 and x

Igoryamba

Answer:

13x

Step-by-step explanation:

The product of 13 and x would be 13*x or just simply 13x.

6 0

3 years ago

Kyle is raising funds for his company by selling preferred stock. The preferred stock has a par value of $122 and a dividend ra

kirza4 [7]

no i think your answer is 9082

6 0

3 years ago

Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

3 years ago