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2 years ago

0.75x+1.5=-3 what’s the answer? tysm for any help! (:

Mathematics

2 answers:

nadezda [96]2 years ago

8 0

The answer is x is -6

lesya692 [45]2 years ago

7 0

Answer:

x=−6

Step-by-step explanation: Brainliest:)

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Captain Salamander has a lazy time on the trip and spends about 15 hours a day sleeping. What fraction of the day does he spend

padilas [110]

Well there is 24 hours in a day so 24 - 15 = 9. So he spends 9 hours awake.

7 0

2 years ago

What is the equation of the line that passes through the point (-6,-6) and has a slope of 2/3?

kotegsom [21]

Answer:

-6=2/3*-6

Step-by-step explanation:

y=mx+b

slope = m b=y intercept

6 0

3 years ago

How wide is the hole at ground level on the drawing

Bumek [7]

Answer:

It’s going by 2

Step-by-step explanation:

If you go up it goes to negative and if you go down it goes to positive

7 0

3 years ago

Read 2 more answers

PLS HELPP Help Solve for x <br>X =

Basile [38]

Answer:

x = 9

Step-by-step explanation:

The big triangle is similar to smaller one so we can use the side lengths' proportion to find x:

7/4 = (x-3)+8/8

7/4 = (x+5)/8 cross multiply expressions

56 = 4x + 20 subtract 20 from both sides

36 = 4x divide both sides by 4

9 = x

4 0

2 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