All categories

2 years ago

Help me please giving more than 5 points!

Mathematics

2 answers:

Scilla [17]2 years ago

5 0

To find the median you need to lay out all the answers by lowest to highest. Then, you’ll need to find the number that’s in the middle and that’s the median. However, if there’s 2 middle numbers you’ll need to add them then divide it by 2 and that’s the median.

k0ka [10]2 years ago

3 0

Answer:

Step-by-step explanation:

Find the the numbers in the middle. Since we have an even number of numbers. what is in-between 80 and 82. <u>81</u>

You might be interested in

Norah wants to download 2 different applications on her phone. 1 used 2.42 megabytes. The other 1 used 1.76. How many total mega

CaHeK987 [17]

I believe you add 2.42 and 1.72

3 0

3 years ago

Read 2 more answers

Elizabeth can jog 1 1/2 miles in 1/2 hour. How many miles can she jog in 1 hour

Nadya [2.5K]

Answer:

3 miles

Step-by-step explanation:

1/2 hour is 30 minutes so 1.5 x 2 is 3

5 0

3 years ago

Read 2 more answers

Write 9 41/80 as a decimal

jolli1 [7]

Leave the whole number alone for now. Divide 41 by 80 and you get 0.5125

So your answer is 9.5125

6 0

3 years ago

Forty percent of households say they would feel secure if they had $50,000 in savings. you randomly select 8 households and ask

Kay [80]

Answer:

Let X be the event of feeling secure after saving $50,000,

Given,

The probability of feeling secure after saving $50,000, p = 40 % = 0.4,

So, the probability of not feeling secure after saving $50,000, q = 1 - p = 0.6,

Since, the binomial distribution formula,

$P(x=r)=^nC_r p^r q^{n-r}$

Where, $^nC_r=\frac{n!}{r!(n-r)!}$

If 8 households choose randomly,

That is, n = 8

(a) the probability of the number that say they would feel secure is exactly 5

$P(X=5)=^8C_5 (0.4)^5 (0.6)^{8-5}$

$=56(0.4)^5 (0.6)^3$

$=0.12386304$

(b) the probability of the number that say they would feel secure is more than five

$P(X>5) = P(X=6)+ P(X=7) + P(X=8)$

$=^8C_6 (0.4)^6 (0.6)^{8-6}+^8C_7 (0.4)^7 (0.6)^{8-7}+^8C_8 (0.4)^8 (0.6)^{8-8}$

$=28(0.4)^6 (0.6)^2 +8(0.4)^7(0.6)+(0.4)^8$

$=0.04980736$

(c) the probability of the number that say they would feel secure is at most five

$P(X\leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$

$=^8C_0 (0.4)^0(0.6)^{8-0}+^8C_1(0.4)^1(0.6)^{8-1}+^8C_2 (0.4)^2 (0.6)^{8-2}+8C_3 (0.4)^3 (0.6)^{8-3}+8C_4 (0.4)^4 (0.6)^{8-4}+8C_5(0.4)^5 (0.6)^{8-5}$