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Nimfa-mama [501]

2 years ago

8

light a flashes every 5 seconds Light b flashes every 6 seconds light c flashes every 7 seconds work out how long it will take f

or all of them to flash

1 answer:

Svet_ta [14]2 years ago

8 0

Answer:

210 seconds

Step-by-step explanation:

LCM of 5,6 and 7

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Find the 4th term in the following sequence (n = 4)

lisabon 2012 [21]

this formula, doesn't rely on a product, relies on a "sum", or is namely an arithmetic sequence, aₙ₋₁ -3 is another way of saying, the value of the previous term minus 3, so it relies on the ordinal value of a term, so is an recursive formula, well, let's get it when n = 4.

$\begin{array}{ccll} order&term&value\\ \cline{1-3} 1&a_1&8\\ 2&a_2&a_{2-1}-3\\ &&a_1-3\\ &&8-3\\ &&5\\ 3&a_3&a_{3-1}-3\\ &&a_2-3\\ &&2\\ 4&a_4&a_{4-1}-3\\ &&a_3-3\\ &&2-3\\ &&-1 \end{array}$

5 0

2 years ago

What is the slope of the line that passes through the points (9,6) and<br> (5, -8)?

LiRa [457]

Y2-y1
--------
x2-x1

-8-6
---------
5-9

-14
-----
-4

= 14/4

= 7/2

4 0

3 years ago

Use the exponential function y=500(.9)^x to find the value of the video game console after 4 years.

zysi [14]

Answer:

328.05 dollars

7 years

Step-by-step explanation:

1.

y=500(.9)^4 =328.05

What is being asked in the problem and what does that mean?

We are asked to the price of the video game after 4 years.

What do I know and what does it mean? What plan am I going to try?

We know the <u>initial price is $500</u>, the <u>value depreciates 10% each year </u>because we have .9 or 90% of the price going into the next year.

- value of the video game after first year 90% of 500 so is 450

- value of the video game after second year 90% of 450 so is 405

-value of the video game after third year 90% of 405 so is 364.5

-value of the video game after <u>fourth year</u> 90% of 364.5 so is 328.05

The plan is to substitute x with 4 and calculate y, y=500(.9)^4

What is your answer and what does it mean?

The answer is $328.05, and it means that the video game that was initially worth $500 it lost its' value by 10 % each of the four years.

2.

y= 500(.9)^x

----------------------------------------------------------------------------

x =8, y= 500(0.9)^8 = 215.234 ≈215.23, this is less than $250

x = 7, y= 500(0.9)^7 = 239.148 ≈239.15, this is less than $250

x =6, y= 500(0.9)^6 = 265.721 ≈ 265.72, this is more than $250

What is being asked in the problem and what does that mean?

We are asked to find the value of x that represents the years such that the value of the console is still under $250.

What do I know and what does it mean? What plan am I going to try?

We know the value of y has to be less that $250, we know the inequality

[500(.9)^x ] < 250, the plan is to try different values for x until we have the maximum value of x that gives us less than 250.

8 0

3 years ago

A gardening club has 36 members,of which 29 are males and the rest are females.What is the ratio of females to males?

Jlenok [28]

7:29
i found the answer by doing 29 minus 36 and the answer is how many females are on the gardening club

6 0

3 years ago

A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su

Sladkaya [172]

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let $\mu_1$ = mean sales per market in Region 1.

$\mu_2$ = mean sales per market in Region 2.

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, $H_A$ : > $\mu_1-\mu_2\neq$ 0 {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about population standard deviations;

T.S. = $\frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+ {\frac{1}{n_2}}} }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean sales in Region 1 = 84

$\bar X_2$ = sample mean sales in Region 2 = 78.3

$s_1$ = sample standard deviation of sales in Region 1 = 6.6

$s_2$ = sample standard deviation of sales in Region 2 = 8.5

$n_1$ = sample of supermarkets from Region 1 = 12

$n_2$ = sample of supermarkets from Region 2 = 17

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times 8.5^{2} }{12+17-2} }$ = 7.782

So, <u><em>the test statistics</em></u> = $\frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }$ ~ $t_2_7$

= 1.943

The value of t-test statistics is 1.943.

Now, at a 0.02 level of significance, the t table gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have<u><em> insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

6 0

3 years ago