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2 years ago

Can someone answer this

Mathematics

1 answer:

Lunna [17]2 years ago

8 0

Answer:

$m = \boxed{1} \: \: c = \boxed{4}$

Step-by-step explanation:

$y = x + 4 \\ equating \: it \: with \\ y = mx + c \\ m = \boxed{1} \: \: c = \boxed{4}$

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If 40% of a number is 56, what was the original number?

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2 years ago

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konstantin123 [22]

The data is missing in the question. The data is provided below :

Document : 1 2 3 4 5 6 7 8

Brand A 17 29 18 14 21 25 22 29

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Solution :

State of the hypothesis of the null hypothesis and alternate hypothesis.

Null hypothesis : $h_A = h_B$

Alternate hypothesis : $h_A > h_B$

These hypothesis is a one tailed test. The null hypothesis will get rejected when the mean difference between the sample means is very small.

Significance level = 0.05

Therefore the standard error is : $SE = \sqrt{(\frac{s^2_1}{n_1})+(\frac{s^2_2}{n_2})}$

= 3.602

And the degree of freedom, DF = 14

$t=\frac{(x_1-x_2)-d}{SE}$

= -1.319

Here, $s_1$ = standard deviation of the sample 1

$s_2$ = standard deviation of the sample 2

$n_1$ = size of the sample 1

$n_2$ = size of the sample 2

$x_1$ = mean of the sample 1

$x_2$ = mean of the sample 2

d = the hypothesis difference between the population mean

The observed difference in a sample means t static of -1.32. From t distribution calculator to determine P( $t \leq -1.32$ ) = 0.1042

Since the P value of 0.1042 is greater than significance level o 0.05, we therefore cannot reject the null hypothesis.

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3 years ago

Graph the linear equation -2x-3y=-7

Nana76 [90]

Answer:

hope this helps !

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2 years ago

A box contains 20 light box of which five or defective it for lightbulbs or pick from the box randomly what's the probability th

Snowcat [4.5K]

Answer:

Step-by-step explanation:

Given:-

- The box has n = 20 light-bulbs

- The number of defective bulbs, d = 5

Find:-

what's the probability that at most two of them are defective

Solution:-

- We will pick 2 bulbs randomly from the box. We need to find the probability that at-most 2 bulbs are defective.

- We will define random variable X : The number of defective bulbs picked.

Such that, P ( X ≤ 2 ) is required!

- We are to make a choice " selection " of no defective light bulb is picked from the 2 bulbs pulled out of the box.

- The number of ways we choose 2 bulbs such that none of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 2:

X = 0 , Number of choices = 15 C r = 15C2 = 105 ways

- The probability of selecting 2 non-defective bulbs:

P ( X = 0 ) = number of choices with no defective / Total choices

= 105 / 20C2 = 105 / 190

= 0.5526

- The number of ways we choose 2 bulbs such that one of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 1 and out of defective n = 5 choose r = 1 defective bulb:

X = 1 , Number of choices = 15 C 1 * 5 C 1 = 15*5 = 75 ways

- The probability of selecting 1 defective bulbs:

P ( X = 1 ) = number of choices with 1 defective / Total choices

= 75 / 20C2 = 75 / 190

= 0.3947

- The number of ways we choose 2 bulbs such that both of them are defective, out of 5 available defective bulbs choose r = 2 defective.

X = 2 , Number of choices = 5 C 2 = 10 ways

- The probability of selecting 2 defective bulbs:

P ( X = 2 ) = number of choices with 2 defective / Total choices

= 10 / 20C2 = 10 / 190

= 0.05263

- Hence,

P ( X ≤ 2 ) = P ( X =0 ) + P ( X = 1 ) + P (X =2)

= 0.5526 + 0.3947 + 0.05263

= 1

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2 years ago