If BC=CD=94 and AD=59, what is AB?

Which statement is true? the equation –3|2x 1.2| = –1 has no solution. the equation 3.5|6x – 2| = 3.5 has one solution. the equa

vampirchik [111]

The statement which is true for the provided reason is the equation 3.5|6x – 2| = 3.5 has one solution.

<h3>How to find one solution of equation?</h3>

If the equation is given a unique and single value of variable after solving then the equation is said one solution equation. Such equation gives exactly one solution.

The three types of solution of an equation are-

One solution-When the provides the singe value of the given variable, we get one or unique solution.
No solution-When the graph of equations is equal than it provides the no solution. When the value, does not equate for the expression, then the expression has no solution.
Infinite many solution-For the same line, the equation has the infinite many solutions.

The first equation given in the problem is,

$-3|2x+ 1.2| = -1$

This equation has one solution. This in not correct option.

The second equation given in the problem is,

$3.5|6x - 2| = 3.5 \\|6x-2|=0\\6x=2\\x=\dfrac{2}{6}\\x=\dfrac{1}{3}$

This equation has one solution. This in correct option.

Hence, the statement which is true for the provided reason is the equation 3.5|6x – 2| = 3.5 has one solution.

Learn more about the solution of the equation here;

brainly.com/question/21283540

7 0

2 years ago

Here are summary statistics for randomly selected weights of newborn girls: nequals=174174, x overbarxequals=30.930.9 hg, seq

MaRussiya [10]

Answer:

The 95% confidence interval would be given by (29.780;32.020)

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=30.9$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=7.5 represent the sample standard deviation

n=174 represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=174-1=173$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,173)".And we see that $t_{\alpha/2}=1.97$ , this value is similar to the obtained with the normal standard distribution since the sample size is large to approximate the t distribution with the normal distribution.