6th grade math help me pleaseeee

ABCD is a parallelogram.<br>Angle CDA is 56°<br>What is the measure of angle ABC?

BartSMP [9]

Answer:

Step-by-step explanation:

angle CDA = angle ABC (opposite angles of a parallelogram are equal)

56 degree = angle ABC

therefore angle ABC is 56 degree.

6 0

2 years ago

One number is 3 less than another number if their sum is 49

Musya8 [376]

One number is 26 the other is 23

x+y=49

Y=x+3,X (now substitute it in the equation)

x+3+x=49

2x+3=49

2x=46. X=23

Y=x+3. Y=23+3 =26 for checking 23+26=49

6 0

1 year ago

A tangent is >insert answer here< to the radius that shares the point of tangency

tatuchka [14]

A tangent line is always perpendicular to the radius that shares the point of tangency. In other words, where your tangent line touches the circle (at only one point), that line is perpendicular to the radius at that point.

3 0

2 years ago

The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.

salantis [7]

a) $X$

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

a)

In this problem, the score on the exam is normally distributed with the following parameters:

$\mu=78.1$ (mean)

$\sigma = 10.8$ (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

$X$

And the probability for this to occur can be written as:

$p(X$

b)

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between $z=-\infty$ and $z=Z$ , where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

$Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02$

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

$p(X$

And by looking at the z-score tables, we find that this probability is:

$p(z$

And so,

$p(X$

c)

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

$p(X>89.1)$

First of all, we have to calculate the z-score corresponding to this value of X, which is:

$Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02$

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

$p(z>1.02) =p(z$