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4 years ago

In a hurry pls help, will give brainliest to the best answer!

2 answers:

7 0

I think it's B. A number line shows numbers from 0 to 12 in increments of 1. The portion of the line from 1 to 7 is bold with shaded circles at 1 and at 7.

Please correct me if I'm wrong!! :)

Andreyy894 years ago

3 0

B) A number line shows numbers from 0 to 12 in increments of 1. The portion of the line from 1 to 7 is bold with shaded circles at 1 and at 7

Hope this Helps!!

You might be interested in

The sum of two polynomials is 8d5 – 3c3d2 + 5c2d3 – 4cd4 + 9. If one addend is 2d5 – c3d2 + 8cd4 + 1, what is the other addend?

lidiya [134]

This one is a bit confusing but I'll try to answer. Since the sum is given and one of the addend is given, we need to perform subtraction to get the other addend.

8d⁵ - 3c³d² + 5c²d³ - 4cd⁴ + 9 ⇒ SUM
<u>- 2d⁵ - c³d² + 8cd⁴ + 1 </u> ⇒ 1ST ADDEND
6d⁵ - 2c³d² + 5c²d³ -12cd⁴ - 8 ⇒ 2ND ADDEND

8 0

3 years ago

Read 2 more answers

What is 0.98989898 with bar notation?

Tom [10]

Answer:

Man it's easy

0.98.... = —–

0.98

Step-by-step explanation:

try putting a line above the 98.

if it helps don't forget to like and markme

4 0

3 years ago

The Center for Medicare and Medical Services reported that there were 295,000 appeals for hospitalization and other Part A Medic

Ymorist [56]

Answer:

(a) 0.00605

(b) 0.0403

(c) 0.9536

(d) 0.98809

Step-by-step explanation:

We are given that 40% of first-round appeals were successful (The Wall Street Journal, October 22, 2012) and suppose ten first-round appeals have just been received by a Medicare appeals office.

This situation can be represented through Binomial distribution as;

$P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....$

where, n = number of trials (samples) taken = 10

r = number of success

p = probability of success which in our question is % of first-round

appeals that were successful, i.e.; 40%

So, here X ~ $Binom(n=10,p=0.40)$

(a) Probability that none of the appeals will be successful = P(X = 0)

P(X = 0) = $\binom{10}{0}0.40^{0}(1-0.40)^{10-0}$

= $1*0.6^{10}$ = 0.00605

(b) Probability that exactly one of the appeals will be successful = P(X = 1)

P(X = 1) = $\binom{10}{1}0.40^{1}(1-0.40)^{10-1}$

= $10*0.4^{1} *0.6^{10-1}$ = 0.0403

P(X >= 2) = 1 - P(X = 0) - P(X = 1)

= 1 - $\binom{10}{0}0.40^{0}(1-0.40)^{10-0} - \binom{10}{1}0.40^{1}(1-0.40)^{10-1}$

= 1 - 0.00605 - 0.0403 = 0.9536

(d) Probability that more than half of the appeals will be successful = P(X > 0.5)

For this probability we will convert our distribution into normal such that;

X ~ N( $\mu = n*p=4,\sigma^{2}= n*p*q = 2.4$ )

and standard normal z has distribution as;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

P(X > 0.5) = P( $\frac{X-\mu}{\sigma}$ > $\frac{0.5-4}{\sqrt{2.4} }$ ) = P(Z > -2.26) = P(Z < 2.26) = 0.98809

3 0

3 years ago

a shipping box will be filled with two different machine parts. one type of machine part weights 0.15 ponds and the other weight

gulaghasi [49]

$\frac{x}{2} =0.15$ hope so this helps

8 0

3 years ago

How would u do this? i needz helpz. plz helpz

castortr0y [4]

You have to add them all together and times that by 60%, when you get 13.8 that's the answer and to prove it, subtract 23-13.8 and you get 9.2 and that's 40%

3 0

3 years ago