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3 years ago

A bat and ball cost $1.10.

Mathematics

2 answers:

ArbitrLikvidat [17]3 years ago

5 0

Answer: The answer is 5 cents

Step-by-step explanation: The reason this answer is 5 cents is because the bat costs one dollar MORE than the ball. Which means that the bat costs $1.05 and the ball costs 5 cents.

pogonyaev3 years ago

4 0

Answer:

$0.10

Step-by-step explanation:

If the total is $1.10, and the bat costs $1 then you would subtract $1 from $1.10 (1.10-1) and get .10 meaning the ball costs 10 cents

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F(x) = x2 – 7, and g(x) = 5 – x2 then g(f(-4)) =

Troyanec [42]

Step-by-step explanation:

find the value of f(-4) first

f(-4) = (-4)^2 - 7 ➡ 9 use this to find g(f(-4))

g(9) = 5 - 9^2 ➡ -76

8 0

3 years ago

Can someone please answer this ASAP?

Ipatiy [6.2K]

Answer:

Letter C

Step-by-step explanation:

Given:

$5a+18$

Subtract 18 from both sides

$5a$

Divide 5 from both sides to get $a$ alone

$a$

Letter C is the correct answer choice because the dot is at -9, the arrow is facing to the left, and the dot is open indicating that it's not greater/less than ""or equal to"".

Hope this is helpful

6 0

3 years ago

A graphic designer chose a base font size and represented it as 1 on the scale. he then listed some consecutive scale sizes. two

garik1379 [7]

Answer:

3.24

Step-by-step explanation:

You would use the two fonts given to find the scale factor.

So the scale factor would be 10.4976/5.832 = 1.8

So if the scale factor is 1.8 we can use that to find the font size just before that. To do this you would divide the font size by the scale factor.

5.832/1.8 = 3.24

The font size just before 5832 is 3.24.

Hope this helps, if there is an error in my work please correct me.

4 0

3 years ago

g On any day, the probability of rain is 0.3. The occurrence of rain on any day is independent of the occurrence of rain on any

Ksju [112]

Answer:

0.249579

Step-by-step explanation:

P(rain) = 0.3

P(no rain) = 1 - 0.3 = 0.7

The event of rain falling a second time within the next 5 days is possible in these ways

1. Rain on days 1 and 2

2. Rain on days 1 and 3; none on day 2

3. Rain on days 1 and 4; none on days 2 and 3

4. Rain on days 1 and 5; none on days 2, 3 and 4

5. Rain on days 1 and 6; none on day 2, 3, 4 and 5

$P(\text{second rain within 5 days}) = 0.3^2+0.3^2\times0.7+0.3^2\times0.7^2+0.3^2\times0.7^3+0.3^2\times0.7^4 = 0.3^2(1+0.7+0.7^2+0.7^3+0.7^4)= 0.09\times2.7731=0.249579$

5 0

3 years ago

Can you please help meeeeeeee

Ahat [919]

The outlier would be 68

4 0

2 years ago