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2 years ago

6

The temperature for 5 days is shown in the table. If on the 6th day the temperature was 83°, by how much will this increase the

overall median temperature?
A) 1°.
B) 2°.
C) 3°.
D) 4°.

1 answer:

Gala2k [10]2 years ago

4 0

It’s b I’m pretty sure sorry if wrong

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uysha [10]

The answer is h! Hope this helps!

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3 years ago

Mimi has 16 more bouncy balls than leah.

Lynna [10]

I would think it would be r

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3 years ago

Read 2 more answers

GEOMETRY URGENT HELP NEEDED

son4ous [18]

First we find the measurement of dark dot .

ANd for that we have to use the formula of the angle formed by the intersection of chord .

Let the dark dot be t.

So we have

$t =\frac{1}{2} (96+66)=81$

t and z are linear pair and sum of angles of linear pair =180 degree.

So we get

$t+z=180 \\ 81+z =180 \\ z = 180-81 =99 degree$

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3 years ago

Explain and demonstrate how to turn 4/20 into a percent

iren2701 [21]

Answer:

20%

Step-by-step explanation:

We can start by looking at it as a ratio, so 4 divided by 20 and 4/20 = 0.2, so that would be 20% viewing 1=100%.

Another way is

20 here is 100%, and for 20 to equal 100, we have to multiply it by 5. To balance it out, multiply the top by 5 as well, and 4 * 5 =20. Therefore, 4/20 = 20%

4 0

3 years ago

The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to rep

Alchen [17]

Ooh, fun

geometric sequences can be represented as
$a_n=a(r)^{n-1}$
so the first 3 terms are
$a_1=a$
$a_2=a(r)$
$a_2=a(r)^2$

the sum is -7/10
$\frac{-7}{10}=a+ar+ar^2$
and their product is -1/125
$\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3$

from the 2nd equation we can take the cube root of both sides to get
$\frac{-1}{5}=ar$
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
$\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r$
subsituting -1/5 for ar
$\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r$
which simplifies to
$\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}$
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for $ax^2+bx+c=0$
$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
so
for 2r²-5r+2=0
a=2
b=-5
c=2

$r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}$
$r=\frac{5 \pm \sqrt{25-16}}{4}$
$r=\frac{5 \pm \sqrt{9}}{4}$
$r=\frac{5 \pm 3}{4}$
so
$r=\frac{5+3}{4}=\frac{8}{4}=2$ or $r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}$

use them to solve for the value of a
$\frac{-1}{5}=ar$
$\frac{-1}{5r}=a$
try for r=2 and 1/2
$a=\frac{-1}{10}$ or $a=\frac{-2}{5}$

test each
for a=-1/10 and r=2
a+ar+ar²= $\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}$
it works

for a=-2/5 and r=1/2
a+ar+ar²= $\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}$
it works

both have the same terms but one is simplified

the 3 numbers are $\frac{-2}{5}$ , $\frac{-1}{5}$ , and $\frac{-1}{10}$

6 0

3 years ago