1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Luden [163]
3 years ago
10

What is the slope of the line between (3,-4) and (-2, 1)? (1 point)

Mathematics
2 answers:
e-lub [12.9K]3 years ago
7 0

Answer:

The slope is -1

Step-by-step explanation:

Slope formula:

-4-1/3-(-2)

-5/5

-1

sveta [45]3 years ago
6 0

Answer:

m = -1

Step-by-step explanation:

You might be interested in
Bob and his friend bill have the same birthday,but bob is 3 years older than bill.let the variable x represent bobs age and y re
worty [1.4K]
Y + x=
are there answer choices  ?
6 0
3 years ago
The z-score associated with 95% is 1.96. If the sample mean is 200 and the standard deviation is 30, find the upper limit of the
ziro4ka [17]

Answer:

The upper limit of the 95% confidence interval is:

C.I_u = 200 + (58.8/\sqrt{n})

Step-by-step explanation:

The formula is given as:

C.I = μ ± Z*σ/\sqrt{n}

The upper limit => C.I_u = μ + Z*σ/\sqrt{n}

The lower limit => C.I_l = μ - Z*σ/\sqrt{n}

The sample size (n) is not stated in the question. Hence, we calculate the upper limit with respect to n.

The upper limit => C.I_u = 200 + 1.96*(30/\sqrt{n})

                                    = 200 + (1.96*30)/\sqrt{n}

                                    = 200 + 58.8/\sqrt{n}

5 0
3 years ago
Help me on this question please​
Luba_88 [7]
Sorry hun I’ve never learned that
5 0
2 years ago
Tina wants to save money for school.Tina invests $1,000 in an account that pays an interest rate of 6.75%. How many years will i
shepuryov [24]
It take 2.75 years
Step by step explanation:
18600$ / 1000$x6.75%=2.75
5 0
3 years ago
In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A
Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let p_1 be the probability that a bearing meets the specification.

So, p_1=0.9

Sample size, n_1=500, is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: \mu_1=n_1p_1=500\times0.9=450

Variance: \sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45

\Rightarrow \sigma_1 =\sqrt{45}=6.71

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, X\geq 440.

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56.

So, the probability that a given shipment is acceptable is

P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062

Hence,  the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by p_2.

p_2=0.94

The total number of shipment, i.e sample size, n_2= 300

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, \mu_2, and variance, \sigma_2^2.

Mean: \mu_2=n_2p_2=300\times0.94=282

Variance: \sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92

\Rightarrow \sigma_2=\sqrt(16.92}=4.11.

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85.

So, the probability that a given shipment is acceptable is

P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977

Hence,  the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, \alpha be the required probability of acceptance of one shipment.

So,

-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}

On solving

\alpha= 0.977896

Again, the probability of acceptance of one shipment, \alpha, depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let p be the probability that one bearing meets the specification. So

-2.01=\frac{439.5-500  p}{\sqrt{500 p(1-p)}}

On solving

p=0.9053

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0
3 years ago
Other questions:
  • A new car is purchased for 15500 dollars. The value of the car depreciates at 7.5% per year. What will the value of the car be,
    10·1 answer
  • A rectangle has an area of k2+19k+60 square inches. If the value of k and the dimensions of the rectangle are all natural number
    13·1 answer
  • There are 62 rows of 9 chairs about how many chairs are there???
    11·1 answer
  • A company is holding a sales contest for its sales reps. 12 sales reps are based in Office A and 7 sales reps are based in Offic
    5·1 answer
  • Davida draws a shape with 5 sides. two sides are each 5 inches long. two other sides are each 4 inches long. the perimeter of th
    15·1 answer
  • Find the equation of a line that passes through A(3, 5) and with slope m=4
    13·2 answers
  • Gordon types 3,572 words in 47 minutes. Find the unit rate.
    15·2 answers
  • Put them in order from the least to greatest. You can use the number line to help!!
    13·1 answer
  • 70% of 17 is what number?
    14·2 answers
  • Four less than twice a number is 16
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!