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3 years ago

Can someone solve this

Mathematics

1 answer:

77julia77 [94]3 years ago

6 0

Answer:

A ≈ 269.81 cm²

Step-by-step explanation:

This shape is a hexagon (it has 6 sides) so let's use the formula for the area of a hexagon.

A = $\frac{3\sqrt{3} }{2} s^{2}$

Where s is the length of the sides

Substitute:

A = $\frac{3\sqrt{3} }{2} s^{2}$

A = $\frac{3\sqrt{3} }{2} 10^{2}$

Solve:

A = $\frac{3\sqrt{3} }{2} 10^{2}$

A = $\frac{3\sqrt{3} }{2} 100$

Multiply the numerator:

A = $\frac{3\sqrt{3} }{2} 100$

A ≈ $\frac{5.12 }{2} 100$

(The numerator reflects a rounded number, but the actual calculations are exact)

Divide the fraction:

A ≈ $\frac{5.12 }{2} 100$

A ≈ 2.6(100)

Multiply:

A ≈ 2.6(100)

A ≈ 269.81 cm²

Therefore, the area is approximately 269.82 cubic centimeters.

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The length of the top of a computer desk is 2 1/4 feet longer than it’s width. If it’s width measures y feet, express its length

Lera25 [3.4K]

Answer:

The length as an algebraic expression in y is $y + (2\frac{1}{4} )$ ft

Step-by-step explanation:

The measure of the width of the computer desk = y feet

The measure of the length of the desk = width + 2 1/4 ft

⇒Measure of Length = $y + (2\frac{1}{4} )$ ft

Hence, the length as an algebraic expression in y is $y + (2\frac{1}{4} )$ ft

5 0

3 years ago

Each flower has a specific price. One tulip, one daisy, and 2 mums cost $4.20. On tulip, two daisies and one mum cost $3.80. Two

sineoko [7]

Answer:

The answer to your question is $5.60

Step-by-step explanation:

1 tulip + 1 daisy + 2 mums = $4.20 (1)

1 tulip + 2 daisies + 1 mum = $ 3.80 (2)

2 tulips + 2 mums = $4.80 (3)

1 tulip + 3 daisies + 2 mums = ?

tulip = t daisy = d mum = m

Process

1.- Find the price of each flower using system of equations

Solve equation 3 for tulip

2t = 4.80 - 2 m divide by 2 (4)

t = 2.40 - 1m

2.- Substitute equation 4 in equation 1 and 2

(2.4 - 1m) + 1 d + 2 m = 4.2

(2.4 - 1m) + 2d + 1 m = 3.8

3.- Simplify and solve for each flower

2.4 - 1m + 1d + 2m = 4.2

1d + 1m = 4.2 - 2.4

1d + 1m = 1.8 Equation (5)

2.4 - 1m + 2d + 1m = 3.8

2d = 3.8 - 2.4

2d = 1.4

d = 1.4/2

1d = 0.7

4.- Substitue d in equation 5

1(0.7) + 1m = 1.8

0.7 + 1m = 1.8

1m = 1.8 - 0.7

1m = 1.1

5.- Subtitute m in equation 4

1t = 2.40 - 1(1.1)

1t = 2.40 - 1.1

1t = 1.3

6.- Result

1 tulip + 3 daisies + 2 mums cost = 1.3 + 3(0.7) + 2(1.1)

= 1.3 + 2.1 + 2.2

= $5.6

8 0

3 years ago

Please help me with this one I beg you!! And don’t send those file links I know that their scams. Please help I will mark brainl

borishaifa [10]

Step-by-step explanation:

2;6 -> 6; -2

4;4 -> 4 ; -4

6;8 -> 8; -6

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2 years ago

I need help with this math question

Oksanka [162]

1.254 × 10 ^11

would be the answer

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2 years ago

Read 2 more answers

The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella germanica. A study investigated

cluponka [151]

Answer:

We conclude that there is no difference in the proportion of cockroaches that died on each surface.

Step-by-step explanation:

We are given that a study investigated the persistence of this pesticide on various types of surfaces.

After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On the glass, 18 cockroaches died, while on plasterboard, 25 died.

Let $p_1$ = proportion of cockroaches that died on glass surface.

$p_2$ = proportion of cockroaches that died on plasterboard surface.

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 {means that there is no difference in the proportion of cockroaches that died on each surface}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq$ 0 {means that there is a significant difference in the proportion of cockroaches that died on each surface}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of cockroaches that died on glass surface = $\frac{18}{36}$ = 0.50

$\hat p_2$ = sample proportion of cockroaches that died on plasterboard surface = $\frac{25}{36}$ = 0.694

$n_1$ = sample of cockroaches on glass surface = 36

$n_2$ = sample of cockroaches on plasterboard surface = 36

So, test statistics = $\frac{(0.50-0.694)-(0)}{\sqrt{\frac{0.50(1-0.50)}{36}+\frac{0.694(1-0.694)}{36} } }$

= -1.712

The value of z test statistics is -1.712.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the proportion of cockroaches that died on each surface.

8 0

3 years ago