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3 years ago

Can someone help me on this. I'm really confused

Mathematics

1 answer:

umka21 [38]3 years ago

5 0

Answer: I believe its 120

Step-by-step explanation:

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A ___ is a transformation that changes the size but not the shape of a figure. A dilation can ____ or ____ a figure

Liula [17]

Answer:

1) Dilation

2) enlarge or shrink

Step-by-step explanation:

1) The shape stays the same, but the area and size change based upon the scale factor (k)

2) Dilation can shrink or enlarge something based upon a scale factor

Vote Brainliest!!

8 0

3 years ago

What is the equivalent fraction Of 10 over 15

Rudiy27

Answer:

2/3

Step-by-step explanation:

4 0

3 years ago

15 points please help :")

Liono4ka [1.6K]

Answer:

15*C

Step-by-step explanation:

The temperature rises by 2.4 every 3 cycles per second. So you would have to keep the pattern going.

6 cycles p/second = 19.8*C

3 cycles p/second = 17.4*C

0 cycles p/second = 15.0*C

So, because 0 cycles means the engine is resting, 15*C would be the answer.

4 0

3 years ago

Read 2 more answers

Complete the point-slope equation of the line through (2,3) and (7,4) <br> y-4=

Anna71 [15]

Answer:

y - 4 = $\frac{1}{5}$ (x - 7)

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (2, 3) and (x₂, y₂ ) = (7, 4)

m = $\frac{4-3}{7-2}$ = $\frac{1}{5}$

Use either of the 2 points for (a, b)

Using (7, 4), then

y - 4 = $\frac{1}{5}$ (x - 7) ← in point- slope form

4 0

3 years ago

The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co

Alona [7]

Answer:

95% Confidence interval for the variance:

$3.6511\leq \sigma^2\leq 34.5972$

95% Confidence interval for the standard deviation:

$1.9108\leq \sigma \leq 5.8819$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

$s^2=2.89^2=8.3521$

The confidence interval for the variance is:

$\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}$

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

$\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128$

Then, the confidence interval can be calculated as:

$\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972$

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

$\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819$

6 0

3 years ago