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3 years ago

Help!!!!!!!!!!!!!!!!

Mathematics

1 answer:

valkas [14]3 years ago

5 0

Under box 36x + 9, Drag the following:

A.) 9(4x + 1)
D.) (9 . 4x) + (9 . 1)

Under box 9(4x - 1), Drag the following:
B.) (3 . 12x) - (3 . 3)
C.) 36x - 9

Under box (4 . 9x) + (4 . 2), Drag the following:
E.) 4(9x + 2)
F.) 36x + 8

Hope this helps!

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Daniela is a stain glass artist; she created the window below for her living room. What is

OLga [1]

The window has a shape of a semi-circle (half of a circle).

We know the diameter d = 12 ft.

The area of a (complete) circle is given by the formula:
A = π × r²

The radius can be found by halving the diameter:
r = d ÷ 2
= 12 ÷ 2
= 6 ft

Therefore the area of the circle will be:
A = π × r²
= 3.14 × 6²
= 3.14 × 36
= 113.04 ft²

And the area of the semi-circle:
A = 113.04 ÷ 2
= 56.52 ft²

Hence, the area of the window will be 56.52 ft².

3 0

3 years ago

A circle has a radius of 1.4 millimeters and a central angle EOG that measures 21°. What is the length of the intercepted arc EG

Dafna11 [192]

Arc length = C*(21/360)
so the answer is D

7 0

3 years ago

Giselle spent 7/10 of her money on a book and 10/100 of her money on food. What fraction of her money did she spend altogether?

aleksklad [387]

Answer:

80/100 or 80%

Step-by-step explanation:

7/10 = 70/100

70/100 + 10/100 = 80/100

8 0

3 years ago

Read 2 more answers

A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Kar

ch4aika [34]

Answer:

We conclude that the mean waiting time is less than 10 minutes.

Step-by-step explanation:

We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.

Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.8 minutes with a standard deviation of 2.5 minutes.

Let $\mu$ = mean waiting time for bus number 14.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 10 minutes {means that the mean waiting time is more than or equal to 10 minutes}

Alternate Hypothesis, $H_A$ : $\mu$ < 10 minutes {means that the mean waiting time is less than 10 minutes}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

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

where, $\bar X$ = sample mean waiting time = 7.8 minutes

s = sample standard deviation = 2.5 minutes

n = sample of different occasions = 18

So, test statistics = $\frac{7.8-10}{\frac{2.5}{\sqrt{18} } }$ ~ $t_1_7$

= -3.734

The value of t test statistics is -3.734.

Now, at 0.01 significance level the t table gives critical value of -2.567 for left-tailed test.

Since our test statistic is less than the critical value of t as -3.734 < -2.567, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean waiting time is less than 10 minutes.

5 0

3 years ago

X and Y are continuous values, both measured to 2 significant figures. X=230 and Y=400. work out the greatest possible value of

kirill115 [55]

"Two significant figures" means that a digit after the second one could actually be anything ... it wasn't measured. 'X' could be anything from 230 to 239. 'Y' could be anything from 400 to 409. The greatest possible value of that fraction is. (239)/(409 squared) .

8 0

3 years ago