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2 years ago

HELPP URGENT!!! NOWW!!!!!!!! 30 POINTS!!!!!!!!!!

Mathematics

1 answer:

vesna_86 [32]2 years ago

6 0

Answer:

y= 6x

Step-by-step explanation:

Slope is 6

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22/30 in lowest terms

Arada [10]

11/15, you would just divide 22/30 in half and that's as far as it can go.

3 0

3 years ago

Read 2 more answers

A simple random sample of the sitting heights of 36 male students has a mean of 92.8cm. The population of males has sitting heig

Naily [24]

Answer:

there is sufficient evidence to support the claim that male student have a sitting height different of 91.4 cm.

Step-by-step explanation:

p value = 0.0198

level of significance = 0.05

hypothesis:-

$H_{0}:\mu=91.4$

$H_{1}:\mu\neq91.4$

decision:-

p value = 0.0198 <0.05

so, we reject the null hypothesis.]

8 0

3 years ago

John and Jenny together have 45 marbles. Both of them lost 5 marbles each and the product of the number of marbles they now have

Tju [1.3M]

Answer:

124÷2=62

Step-by-step explanation:

Each person has 62 marbles.

The beginning of the problem is not needed to answer the question. The only thing you need to know to answer the question is that they have 124 marbles together.

3 0

3 years ago

. The dot plot shows the number of hours students spent last week on social networking sites. Which is NOT an inference that can

Orlov [11]

B. Females typically spend less time than males on the sites

8 0

2 years ago

. Among all the income tax forms filed in a certain year, the mean tax paid was $2000, and the standard deviation was $500. In a

Anarel [89]

Answer:

a) 0.8413

b) 0.6293

Step-by-step explanation:

Data provided:

Mean tax paid = $2000

Standard deviation = $500

Sample size, n = 625

n=625

Now,

a) P( average tax paid on the sample forms is greater than $1980)

⇒ P(X > 1980)

⇒ $P(\frac{(X-mean)}{\frac{s}{\sqrt n}})$

⇒ $P(\frac{(1980-2000)}{\frac{500}{\sqrt{625}}})$

⇒ P(Z > -1)

= 0.8413 (From standard normal table)

b) P(more than 60 of the sampled forms have a tax of greater than $3000)

given: p = 10% = 0.1

Now, by using CLT

mean = n × p

= 625 × 0.1

= 62.5

Standard deviation, s = $\sqrt{n\times p\times(1-p)}$

= $\sqrt{625\times0.1\times(1-0.1)}$

= 7.5

Thus,

P(X > 60)

= $P(\frac{(X-mean)}{s} > \frac{(60-62.5)}{7.5})$

= P(Z > -0.33)

= 0.6293 (From standard normal table)

4 0

3 years ago