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3 years ago

At 7:00 a.m., the temperature was -3°F. At 6:00 p.m., the temperature was

Mathematics

1 answer:

san4es73 [151]3 years ago

4 0

Answer:

Step-by-step explanation:

We are given the following information: At 7:00 a.m., the temperature was -3°F. At 6:00 p.m., the temperature was 9°F greater than the temperature at 7:00 a.m. This means the temperature at 6:00 p.m. is calculated as -3 + 9.

We know this last temperature is 4°F less than the temperature at noon, thus the temperature at noon is calculated as -3 + 9 + 4

A) -3 + 9 + 4

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Last week, the members of a weight loss support group collectively weighed 1,580 pounds. They just weighed in again and are now

rosijanka [135]

Answer: 1,501

Step-by-step explanation:

100% = 1,580

100% - 5% lost

95%

1580 x .95 = 1501

3 0

3 years ago

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

3 years ago

WILL MARK BRAINLIEST

sp2606 [1]

To make this new triangle I am going to pretend that the original triangle was at the points (4,4) (-4,4) and (0,8). So if this was dilated down by 1/2 then the triangles points would be (2,2) (-2,2) and (0,4). Then if it is reflected over the x-axis the new points would be (2,-2) (-2,-2) and (0,-4) and if it was translated left by 2 units then the points would be (0,-2) (-4,-2) and (-2, -4), then if the triangle was translated up by 4 units the points and the new triangle would be (0,2) (-4,2) and (-2,0). And that would be what the new triangle would be.

The triangle went from (4,4) (-4,4) and (0,8) to (0,2) (-4,2) and (-2,0).

7 0

4 years ago

(-1).(2-6)^2÷8+8-3.4

Paha777 [63]

(-1)*(2-6)^2÷8+8-3.4

(-1)*(-4)^2÷8+8-3.4

-1*16÷8+8-3.4

-16÷8+8-3.4

-2+8-3.4

2.6

6 0

3 years ago

An oak tree is 120 feet taller than the adjacent house. If the tree is also two and a half times the height of the house, how ta

Harman [31]

Let height of house be x.

Oak tree = 120 + x

Also given,

Oak tree = 2.5x

120 + x = 2.5x

1.5x = 120

0.3x = 24

3x = 240

x = 80

Hence, the height of the house is 80 ft.

3 0

3 years ago