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2 years ago

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21% of the students at a High are in Maths group, 15 % are in science group, and 6 % are in both. If a student is selected rando

mly , what is the probability that the student is in maths or science group?

1 answer:

mihalych1998 [28]2 years ago

4 0

Answer:

6%

Step-by-step explanation:

The probability of picking a maths student = 21% = 21/100

The probability of picking a science student = 15% = 15/100 = 3/20

The probability of picking in both = 6/100 = 3/50

The probability that the student is in maths or science = 21/100 - 15/100 = 6/100 = 6%.

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Resistors for electronic circuits are manufactured on a high-speed automated machine. The machine is set up to produce a large r

Anettt [7]

Answer:

See explanation

Step-by-step explanation:

Given

See attachment for proper presentation of question

Required

Mean and Range

To do this, we simply calculate the mean and the range of each row.

$\bar x = \frac{\sum x}{n}$ ---- mean

Where:

$n = 4$ ---- number of rows

$R = Highest - Lowest$ --- range

So, we have:

Sample 1

$\bar x_1 = \frac{1027+ 994 +977 +994 }{4}$

$\bar x_1 = 998$

$R_1 = 1027- 994$

$R_1 = 33$

Sample 2

$\bar x_2 = \frac{975 +1013 +999 +1017}{4}$

$\bar x_2 = 1001$

$R_2 = 1017 - 975$

$R_2 = 42$

Sample 3

$\bar x_3 = \frac{988 +1016 +974 +997}{4}$

$\bar x_3 = 993.75$

$R_3 = 1016-974$

$R_3 = 42$

Sample 4

$\bar x_4 = \frac{998 +1024 +1006 +1010}{4}$

$\bar x_4 = 1009.5$

$R_4 = 1024 -998$

$R_4 = 26$

Sample 5

$\bar x_5 = \frac{990 +1012 +990 +1000}{4}$

$\bar x_5 = 998$

$R_5 = 1012 -990$

$R_5 = 22$

Sample 6

$\bar x_6= \frac{1016 + 998 +1001 +1030}{4}$

$\bar x_6= 1011.25$

$R_6= 1030-998$

$R_6= 32$

Sample 7

$\bar x_7 = \frac{1000 +983 +979 +971}{4}$

$\bar x_7 = 983.25$

$R_7 = 1000-971$

$R_7 = 29$

Sample 8

$\bar x_8 = \frac{973 +982 +975 +1030}{4}$

$\bar x_8 = 990$

$R_8 = 1030-973$

$R_8 = 57$

Sample 9

$\bar x_9 = \frac{992 +1028 +991 +998}{4}$

$\bar x_9 = 1002.25$

$R_9 = 1028 -991$

$R_9 = 37$

Sample 10

$\bar x_{10} = \frac{997 +1026 +972 +1021}{4}$

$\bar x_{10} = 1004$

$R_{10} = 1026 -972$

$R_{10} = 54$

Sample 11

$\bar x_{11} = \frac{990 +1021 +1028 +992}{4}$

$\bar x_{11} = 1007.75$

$R_{11} = 1028 -990$

$R_{11} = 38$

Sample 12

$\bar x_{12} = \frac{1021 +998 +996 +970}{4}$

$\bar x_{12} = 996.25$

$R_{12} = 1021 -970$

$R_{12} = 51$

Sample 13

$\bar x_{13} = \frac{1027 +993 +996 +996}{4}$

$\bar x_{13} = 1003$

$R_{13} =1027 -993$

$R_{13} =34$

Sample 14

$\bar x_{14} = \frac{1022 +981 +1014 +983}{4}$

$\bar x_{14} = 1000$

$R_{14} = 1022 -981$

$R_{14} = 41$

Sample 15

$\bar x_{15} = \frac{977 +993 +986 +983}{4}$

$\bar x_{15} = 984.75$

$R_{15} = 993-977$

$R_{15} = 16$

8 0

2 years ago

The board of directors of a company knows that the probability that the carbon emission from the company’s manufacturing unit is

enyata [817]

the company's production unit has a risk of carbon emission

the probability that the company is exceeding the permissible level is 35% =0.35

so there can only be two possibilities, either the company is exceeding the permissible level or its within the permissible level.

and the accuracy of the test that tests the emission level is 85%

so the probability that the carbon emission is within the permissible level is

1 - 0.35 = 0.65

probability that the test measures this accurately is 0.85

therefore when checking the probability that both events happen we have to multiply the 2 probabilities

the probability that carbon emission is within the permissible level and test predicts this outcome is - 0.65 x 0.85 = 0.5525

answer is D.0.5525

6 0

2 years ago

Read 2 more answers

Evaluate the expression when x=-4.<br> x2 +7x-6

ss7ja [257]

•(-4)^2 + 7(-4) -6
• -18

7 0

2 years ago

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I really don’t know how to do this can someone help

Scorpion4ik [409]

Answer:

0.57

Step-by-step explanation:

6 0

3 years ago

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Simplify <br> 7.2x10^3/1.8x10^7

spayn [35]

To simplify and solve this equation we must follow PEMDAS.

P=Parentheses

E=Exponents

M=Multiplication (from left to right)

D=Division (from left to right)

A=Addition

S=Substraction

We begin the process.

10^3=1000

10^7=10000000

New equation looks like this:

7.2*1000/1.8*10000000

We then multiply

7.2*1000= 7200

7200/1.8=4000

4000*10000000= 40000000000

<em>So our final answer is 40000000000.</em>

Hope this helps!

P.S. I forgot this was scientific notation and I solved it a different way. But both ways give the same answer. Let me know if you want the other method.

3 0

3 years ago