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2 years ago

Need help. Can someone help me???

Mathematics

1 answer:

Natali5045456 [20]2 years ago

8 0

Y= 1/2x + 12y= 1/2x + 12

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What are 5 fractions that can simplify to 2/5

enot [183]

4/10
8/20
16/40
32/80
64/160

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3 years ago

Read 2 more answers

Factorise this : 2p²-p-10

gizmo_the_mogwai [7]

Answer:

Step-by-step explanation:

p(2p-1)-10

8 0

2 years ago

SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You are interested in estimating the average SA

kotegsom [21]

Answer:

I should use at least 304 students

Step-by-step explanation:

Margin error (E) = t × sd/√n

E = 40

sd = 300

confidence level (C) = 98% = 0.98

significance level = 1 - C = 1 - 0.98 = 0.02 = 2%

t-value corresponding to 2% significance level and infinity degree of freedom is 2.326

n = (t×sd/E)^2 = (2.326×300/40)^2 = 17.445^2 = 304 (to the nearest integer)

3 0

2 years ago

The graph shows the solution to which system of inequalities?

postnew [5]

Answer:

is A B and D

Step-by-step explanation:

5 0

2 years ago

Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.

lesya [120]

Answer:

0.1994 is the required probability.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 166 pounds

Standard Deviation, σ = 5.3 pounds

Sample size, n = 20

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{5.3}{\sqrt{20}} = 1.1851$

P(sample of 20 boxers is more than 167 pounds)

$P( x > 167) = P( z > \displaystyle\frac{167 - 166}{1.1851}) = P(z > 0.8438)$

$= 1 - P(z \leq 0.8438)$

Calculation the value from standard normal z table, we have,

$P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%$

0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

3 0

2 years ago