All categories

3 years ago

A is the midpoint of BC. BA = 3x+8 AC=23. solve for x

Mathematics

2 answers:

AnnyKZ [126]3 years ago

7 0

Since A is the midpoint of BC, we can assume BA and AC are equal to each other.

We also know that BC = BA + AC = 23 + 23 = 46

Nastasia [14]3 years ago

3 0

Since A is the midpoint of BC, A divides BA and AC equally.
Therefore, BA=AC.
Since BA=AC,
3x+8=23
Or, 3x= 23-8
Or, 3x= 15
Or, x=15/3
Therefore, x=5.
Ans: x=5.

You might be interested in

Ik rp because the i forgot the sc

Stels [109]

Answer:

Step-by-step explanation:

This is a geometric mean problem...a ratio. The geometric mean is the height of the triangle, 6. To solve:

$\frac{x}{6} =\frac{6}{3}$ and cross multiply to solve for x:

3x = 36 so

x = 12

6 0

1 year ago

Read 2 more answers

Express 80 miles per hour in kilometers per hour

Anna [14]

80 miles per hour is 128.74752 kilometers per hour

8 0

3 years ago

Read 2 more answers

Given that

mina [271]

Answer:

huh?

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Weights of the vegetables in a field are normally distributed. From a sample Carl Cornfield determines the mean weight of a box

pantera1 [17]

To find the z-score for a weight of 196 oz., use

$z=\frac{x-\mu}{\sigma}=\frac{196-180}{8}=\frac{16}{8}=2$

A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.

1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.

7 0

3 years ago

You are testing the claim that the mean GPA of night students is different from the mea GPA of day students. You sample 20 night

svetlana [45]

Answer:

As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.

Step-by-step explanation:

Here n= 20

Sample mean GPA = x`= 2.84

Standard mean GPA = u= 2.55

Standard deviation = s= 0.45.

Level of Significance.= ∝ = 0.01

The hypothesis are formulated as

H0: u1=u2 i.e the GPA of night students is same as the mean GPA of day students

against the claim

Ha: u1≠u2

i.e the GPA of night students is different from the mea GPA of day students

For two tailed test the critical value is z ≥ z∝/2= ± 2.58

The test statistic

Z= x`-u/s/√n

z= 2.84-2.55/0.45/√20

z= 0.1441

As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.

8 0

2 years ago