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scoray [572]
3 years ago
7

Which number is equal to 7/10 A. 7 B . 70 C. 0.7 D. 0.07 10 pts

Mathematics
2 answers:
raketka [301]3 years ago
8 0
The right answer is C: 0.7
lutik1710 [3]3 years ago
8 0
B and C because .70 is 70% of 100
You might be interested in
Choose the correct slope of the line that passes through the points (1, −3) and (3, −5).
Flura [38]
Slope of a line passing though point (x1,y1) nd (x2,y2) is
(y2-y1)/(x2-x1)

we have
(1,-3) and (3,-5)
x1=1
y1=-3
x2=3
y2=-5

slope=(-5-(-3))/(3-1)=(-5+3)/(2)=-2/2=-1

slope=-1
4 0
3 years ago
Read 2 more answers
cynthia invests some money in a bank which pays 5% compound interest per year. She wants it to be worth over £8000 at the end of
exis [7]

Answer:

The smallest amount is \£6,911

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

t=3\ years\\ A=\£8,000\\ r=5\%=5/100=0.05\\n=1  

substitute in the formula above

8,000=P(1+\frac{0.05}{1})^{1*3}  

solve for P

8,000=P(1.05)^{3}  

P=8,000/(1.05)^{3}  

P=\£6,910.70  

Round to the nearest pound she can invest

The smallest amount is \£6,911

6 0
3 years ago
A fair die is rolled 12 times. the number of times an even number occurs on the 12 rolls has
bonufazy [111]

Answer:

Step-by-step explanation:

For a fair die, there are six likely options; 1, 2, 3, 4, 5, and 6

the probability of a even number is 3/6 = 0.5

Since the results of the die roll is independent and each trial is mutually exclusive, the distribution to explain the probability of occurrence will follow a binomial distribution such that n is the number of trials

x = number of successful throws

therefore for a Binomial distribution where

P(X =x) = nCx . P^x . (1-P)^ (n-x)

since p = 0.5, and n = 12, the distribution follows

P(X = x) = 12Cx . 0.5^x . (1 - 0.5)^(12- x)

= 12Cx . 0.5^x . 0.5)^(12- x)

where x = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

since we are interested in the probability of the number of times an even number occurs

it can occur either as P(X = 0), P(X =1), P(X =2), P(X =3), P(X =4), P(X =5), P(X =6), P(X =7), P(X =8), P(X =9), P(X =10), P(X =11), and P(X =12)

For no even number in 12 rolls,

P(X = 0) = 12C0 . 0.5^0 . 0.5^(12- 0) = 0.000244

For one even number in 12 rolls,

P(X = 1) = 12C1 . 0.5^1 . 0.5^(12- 1) = 0.002930

For two even number in 12 rolls,

P(X = 2) = 12C2 . 0.5^2 . 0.5^(12- 2) = 0.016113  

For three even number in 12 rolls,

P(X = 3) = 12C3 . 0.5^3 . 0.5^(12- 3) = 0.053711  

For four even number in 12 rolls,

P(X = 4) = 12C4 . 0.5^4 . 0.5^(12- 4) = 0.120850

For five even number in 12 rolls,

P(X = 5) = 12C5 . 0.5^5 . 0.5^(12- 5) = 0.193359

For six even number in 12 rolls,

P(X = 6) = 12C6 . 0.5^6 . 0.5^(12- 6) = 0.225586

For seven even number in 12 rolls,

P(X = 7) = 12C7 . 0.5^7 . 0.5^(12- 7) = 0.193359

For eight even number in 12 rolls,

P(X = 8) = 12C8 . 0.5^8 . 0.5^(12- 8) = 0.120850

For nine even number in 12 rolls,

P(X = 9) = 12C9 . 0.5^9 . 0.5^(12- 9) = 0.053711

For ten even number in 12 rolls,

P(X = 10) = 12C10 . 0.5^10 . 0.5^(12- 10) = 0.016113

For eleven even number in 12 rolls,

P(X = 11) = 12C11 . 0.5^11 . 0.5^(12- 11) = 0.002930

For twelve even number in 12 rolls,

P(X = 12) = 12C12 . 0.5^12 . 0.5^(12- 12) = 0.000244

Final test summation[P(X)] =  1

i.e.

P(X = 0) + P(X =1) + P(X =2) + P(X =3) + P(X =4) + P(X =5) + P(X =6) + P(X =7) + P(X =8) + P(X =9) + P(X =10) + P(X =11) + P(X =12) = 1

Hence since 0.000244 + 0.002930 + 0.016113 + 0.053711 + 0.120850 + 0.193359 + 0.225586 + 0.193359 + 0.120850 + 0.053711 + 0.016113 + 0.002930 + 0.000244 = 1.000000,

the probability value stands

7 0
3 years ago
Please help ill give brainliest
Olenka [21]
1. y=7/8x +1
2. y=-10x
3. y=2.5x-7
4. y=1.2x
5. y=-5x-8
6 0
2 years ago
The atmospheric pressure on an object decreases as altitude increases. If a is the height (in km) above sea level, then the pres
seraphim [82]

Answer:

373.8mmHg

Step-by-step explanation:

a =height (in km) above sea level,

the pressure P(a) (in mmHg) is approximated given as

P(a) = 760e–0.13a .

To determine the atmospheric pressure at 5.458 km, then we will input into the equation

P(5.458km) = 760e–0.13a .

= 760e^(-0.13×5.458)

=760e^-(0.70954)

= 760×0.4919

=373.8mmHg

Therefore, the atmospheric pressure at 5.458 km is 373.8mmHg

3 0
2 years ago
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