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2 years ago

What is the probability that a 51% free-throw shooter will miss her next free throw?

1 answer:

6 0

Assuming all throws are independent of each other, the probability of making the next throw is still 51%, and ... missing it is 100-51=49%

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You purchase 2 action figures per week. You have 48 action figures, which is 75% of the total number of action figures you need

LiRa [457]

Answer:

Step-by-step explanation:

Let the total number of action figures = x

75% of x = 48

$\dfrac{75}{100}*x=48\\\\Multiply \ by \ 100\\\\75*x =48*100\\\\Divide \ by \ 75\\\\x=\dfrac{48*100}{75}\\\\\\x=16*4\\\\x= 64$

Remaining action figures need to be collected = total figures - number of action figures I have

= 64 - 48 = 16

Number of weeks = 16 ÷ 2 = 8

It will take 8 weeks to collect 16 action figures.

3 0

2 years ago

Please help me with this problem! Picture Included!!

alisha [4.7K]

Answer:

y < 1

Step-by-step explanation:

graphing with inequality

6 0

2 years ago

Read 2 more answers

Pls help I’ll brainlest solve it like this

sergey [27]

$\frac{3}{5} t = 6$

$t = 6 \div \frac{3}{5}$

$t = 6 \times \frac{5}{3}$

$t = 2 \times 5$

$t = 10$

____________________________

$\frac{1}{4} a = \frac{14}{15}$

$a = \frac{14}{15} \div \frac{1}{4}$

$a = \frac{14}{15} \times 4$

$a = \frac{56}{15}$

7 0

2 years ago

Read 2 more answers

A coyote had 10 pups in her third litter. It was twice as many as in her second litter, her second litter, and her second litter

Yuki888 [10]

A coyote had 10 pups in her third litter. (10)
It was twice as many as in her second litter,( 2( second litter) = third litter

her second litter had 3 more pups than her first.( first litter +3)=2nd litter

Find the number of pups in her litter

6 0

3 years ago

Write parametric equations of the line through the points (7,1,-5) and (3,4,-2). please use the first point as your base-point w

Roman55 [17]

Given:

A line through the points (7,1,-5) and (3,4,-2).

To find:

The parametric equations of the line.

Solution:

Direction vector for the points (7,1,-5) and (3,4,-2) is

$\vec {v}=\left$

Now, the perimetric equations for initial point $(x_0,y_0,z_0)$ with direction vector $\vec{v}=\left$ , are

$x=x_0+at$

$y=y_0+bt$

$z=z_0+ct$

The initial point is (7,1,-5) and direction vector is $\vec {v}=\left$ . So the perimetric equations are

$x=(7)+(-4)t$

$x=7-4t$

Similarly,

$y=1+3t$

$z=-5+3t$

Therefore, the required perimetric equations are $x=7-4t, y=1+3t$ and $z=-5+3t$ .

5 0

3 years ago