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3 years ago

Triangle after 90 degree rotation

Mathematics

1 answer:

Leno4ka [110]3 years ago

4 0

The answer is D bc you rotate it clockwise one time

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Scott drove 355 miles using 17 gallons of gas. At this rate, how many gallons of gas would he need to drive 284 miles?

Lisa [10]

It should be somewhere around 13.6

3 0

3 years ago

50PTS! Describe the vertical asymptotes and holes for the graph of y = x-5/x^2 -1

Diano4ka-milaya [45]

Ok

first of all, for q(x)/p(x)
if the degree of q(x) is less than the degree of p(x),then the horizontal assemtote is 0

then simplify
any factors you factored out is now a hole, remember them

to find the vertical assemtotes of a function, set the SIMPLIFIED denomenator equal to 0 and solve

so

y=(x-5)/(x^2-1)
q(x)<p(x)
horizontal assemtote is y=0

no factors to simplify so no holes

set denomenator to 0 to find vertical assemtote
x^2-1=0
(x-1)(x+1)=0
x-1=0
x=1

x+1=0
x=-1

the horizontal assemtotes are x=1 and -1

3 0

3 years ago

Read 2 more answers

Guess how old I am. hints: I'm not really in college, I'm in Middle School. When you find out my age you are in for a shock for

babunello [35]

Maybe 11 years old??

4 0

3 years ago

While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem

Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = population proportion.

So, Null Hypothesis, $H_0$ : p = 0.013 {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, $H_A$ : p > 0.013 {means that this is an unusually high number of faulty modems}

The test statistics that would be used here One-sample z-test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion faulty modems= $\frac{10}{367}$ = 0.027

n = sample of modems = 367

So, the test statistics = $\frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }$

= 2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0

3 years ago

Image above!

Rainbow [258]

The answer is C! the crossing of lines is the same on both sides

6 0

3 years ago