All categories

3 years ago

I need help with this someone help me it’s geometry

Mathematics

2 answers:

Nata [24]3 years ago

6 0

Answer:

Step-by-step explanation:

BD is the perpendicular bisector of AC (AB = BC, and ∠CBD = 90°).

Therefore, by side-angle-side congruence (SAS), ΔABD and ΔCBD are congruent.

So AD = CD = 32.

Sindrei [870]3 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

Which form of a quadratic equation is best for finding the y-intercept

Alinara [238K]

Answer:

Step-by-step explanation:

5 0

3 years ago

a sound wave travels through iron at a rate of 5120m/s. at what rate does the sound wave travels in km/h?

Sladkaya [172]

5120 m/s = 18.432<span> kph

</span>Conversion base : 1 mps = 3.6 kphConversion base : 1 kph = 0.27777777777778 mps<span>
</span>

3 0

3 years ago

Please help me with this problem

Burka [1]

Answer:

(x+11)(x-11) its a difference of squares formula, when you FOIL it out the positive 11x and negative 11x will cancel out

Step-by-step explanation:

6 0

3 years ago

Select the correct answer. The radius of a nitrogen atom is 5. 6 × 10-11 meters, and the radius of a beryllium atom is 1. 12 × 1

Grace [21]

A negative exponent means the number of the exponent divide by the number. The the radius of beryllium atom is larger than the radius of a nitrogen atom by two times.

Given information-

The radius of a nitrogen atom is $5.6\times10^{-11}$ meters.

The radius of a beryllium atom is $1.12 \times10^{-10}$ meters

<h3>Negative exponents</h3>

A negative exponent means the number of the exponent divide by the number.

As the radius of the nitrogen atom is $5.6\times10^{-11}$ meters. Convert it in the power of negative exponent of 10.

Let the radius of the nitrogen is $r$ mm. Thus,

$r=5.6\times 10^{-11}\\
r=5.6\times 10^{-10-1}\\
r=5.6\times 10^{-10}\times10^{-1}\\
r=\dfrac{5.6\times 10^{-10}}{10} \\
r=0.56\times 10^{-10}$

Now the base of the exponents of both the atoms are same. As the radius of the nitrogen atom is $0.56\times 10^{-10}$ meters and the radius of the beryllium atom is $1.12 \times10^{-10}$ . Thus the radius of beryllium atom is larger.

To calculate how many times, the radius of beryllium atom should be divided by the radius of a nitrogen atom. Thus,

$=\dfrac{1.12\times10^{-10}}{0.56\times10^{-10}} \\
=2$

Thus the the radius of beryllium atom is larger than the radius of a nitrogen atom by two times.

Learn more about the negative exponents here;

https://brainly.in/question/1221452

7 0

3 years ago

A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of

notsponge [240]

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.01 = 0.99$ , so Z = 2.327.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031$

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.0001 = 2.327\frac{0.0003}{\sqrt{n}}$

$0.0001\sqrt{n} = 2.327*0.0003$

$\sqrt{n} = \frac{2.327*0.0003}{0.0001}$

$(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2$

$n = 48.73$

Rounding up

49 measurements are needed.

7 0

3 years ago

￼ I need help with this someone help me it’s geometry

I need help with this someone help me it’s geometry