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mina [271]
3 years ago
15

8. Given triangle AXTB with x = 8, t = 11 and b = 14, what is the mZT?

Mathematics
1 answer:
Ber [7]3 years ago
5 0
I need the answer aswell
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ATCs are required to undergo periodic random drug testing.
aleksandr82 [10.1K]

Answer:

The responsess to the given question can be defined as follows:

Step-by-step explanation:

P(medication\ used) = 0.007 \\\\P( drug \not\ used ) = 1 - 0.007 = 0.993\\\\P(test\ positive  | medication \ used) = 0.96 \\\\ P(test \ negative | medication \ used) = 1 - 0.96 = 0.04\\\\

P(Point | Non-used\ medication )=0.93\\\\ P(Point |No\ medication ) = 1 - 0.93 = 0.07

As per the given info we draw the tree diagram which is defined in the attachment file.

P ( test \ is\  positive ) = P( medication \ used ) \times P( Test \ positive | medication \ used ) + P( medication not \ used ) \times  P( Test\ positive | medication \ not \ used )

                  = ( 0.007 \times 0.96 ) + ( 0.993\times 0.07 )\\\\= 0.0762

P( medication \ used | test\ positive )= \frac{\textup{P( medication used ) *P( Test positive given medication  used )}}{\textup{P( medication used )}}  

= \frac{( 0.007\times 0.96 )}{0.0762}\\\\= 0.0882

8 0
3 years ago
On saturday, you bowl at mar vista bowl, where renting shoes costs $2 and each game bowled is $3.50. on sunday, you bowl at pinz
ExtremeBDS [4]

Given, at Mar Vista bowl the cost of renting shoes = $2.

The cost for each game bowled = $3.50

There are g games given. So the cost for g games bowled = $(3.50)(g) = $(3.50g)

The total amount soent at Mar Vista Bowl = $(2+3.50g)

At Pinz, the cost for shoe rental = $5.

The cost for each game bowled = $3.25.

So the cost for g game bowled = $(3.25)(g) = $(3.25g)

Total amount spent at Pinz = $(5+3.25g)

Given, the amount spent ateach of the places is same.

so we can write the equation as,

2+3.50g = 5+3.25g

We will move 3.25 to the left side by subtracting it from both sides. we will get,

2+3.50g-3.25g = 5+3.25g-3.25g

2+0.25g = 5

Now we will move 2 to the right side by subtracting 2 from both sides. We will get,

2+0.25g-2 = 5-2

0.25g = 3

Now to get g we will divide 0.25 to both sides. We will get,

\frac{0.25g}{0.25}= \frac{3}{0.25}

g =\frac{3}{0.25}

g = 12

So we have got the required answer.

The number of games bowled = 12.

6 0
3 years ago
Please help!! asap!!! ( 10 points)
MAXImum [283]

Answer:

......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

Step-by-step explanation:

5 0
3 years ago
Please help me solve this
Oduvanchick [21]
\dfrac{x}{ - 3} < - 4 \\\\ x > 12

P. S. The inequality sign changed or flipped when there's a negative figure involved.

Hope this helps. - M
3 0
3 years ago
In circle S with m ∠ R S T = 98 m∠RST=98 and R S = 4 RS=4 units, find the length of arc RT. Round to the nearest hundredth.
user100 [1]

Answer:6.84

Step-by-step explanation:

:)

7 0
3 years ago
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