All categories

3 years ago

Show your work and plz help me !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

AveGali [126]3 years ago

8 0

Answer:

p = 16

Step-by-step explanation:

Since the triangles are similar then the ratios of corresponding sides are equal, that is

$\frac{FD}{HF}$ = $\frac{DE}{FG}$ , substitute values

$\frac{p}{12}$ = $\frac{32}{24}$ ( cross- multiply )

24p = 384 ( divide both sides by 24 )

p = 16

You might be interested in

Guys please help me am I right

Elodia [21]

Answer:

$y \leqslant 6$

Step-by-step explanation:

The range is set of all y-values. The range starts from minimum value to maximum value.

We don't have minimum value (approaching negative infinity.)

We have maximum value at x ≈ 2 equal 6.

Therefore the answer is

-inf <= y <= 6. But we don't usually write that. Instead, we cut out -inf and we get —

$y \leqslant 6$

8 0

2 years ago

Find the distance between the two points in simplest radical form. (3,8) and (5, -1)

ruslelena [56]

Answer: irk

Step-by-step explanation:

7 0

2 years ago

A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq

Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

$s=16.6$ represent the sample deviation

$\bar X=12.8$ represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

$\alpha=1-0.99=0.01$ represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

$\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.