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2 years ago

Workout the value for 3x^2-5y

1 answer:

6 0

Remember to do PEMDAS so your answer is not correct, since you multiplied before doing the parentheses/exponents.

Explanation:
3(3)^2- 5(4)
= 3(9) - 5(4)
= 27 - 20 = 7

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Philip walked a total of 8 kilometers by making 2 trips to school. After 3 trips to school, how many kilometers will Philip have

vekshin1

So one trip = 4 km
two trips = 8
8 + 4 = 12
i believe your answer is 12
Thanks!

7 0

3 years ago

Read 2 more answers

The eye of a dinosaur has 9 yellow, 2 white, 4 black,

vivado [14]

20:5
The total number of jelly beans is 20, and there are 5 orange jelly beans. This is a total to part ratio because it’s comparing the entirety to a part.

8 0

4 years ago

PLEASE HELP MEEEEEEEE AND THANKS

VashaNatasha [74]

Answer:

See below

Step-by-step explanation:

Use 5 for "x" or "a number" and see if the inequality is true.

a) =3 < 5+11 < 2 . . . . false. 16 is not less than 2

b) 7·5 ≥ 35 or 4·5 < -12 . . . . true. 35 = 35

c) -7 + 2·5 = 3 is at most 2 . . . . false. 3 > 2

d) 5·5 +3 = 28 is between 2 and 30 . . . . true.

3 0

3 years ago

A survey of 700 adults from a certain region asked, "What do you buy from your mobile device?" The results indicated that 59%

Kryger [21]

Answer:

a) the test statistic z = 1.891

the null hypothesis accepted at 95% level of significance

b) the critical values of 95% level of significance is zα =1.96

c) 95% of confidence intervals are (0.523 ,0.596)

Step-by-step explanation:

A survey of 700 adults from a certain region

Given sample sizes $n_{1} = 400 and n_{2} = 300$

Proportion of mean $p_{1} = \frac{236}{400} = 0.59 and p_{2} = \frac{156}{300} = 0.52$

Null hypothesis H0 : assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

Alternative hypothesis H1:- p1 ≠ p2

a) The test statistic is

$Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )} } }$

where $p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}$

on calculation we get p = 0.56

now q =1-p = 1-0.56=0.44

$Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )} } }\\ =\frac{0.56-0.52}{\sqrt{0.56X0.44}(\frac{1}{400}+\frac{1}{300} }$

after calculation we get z = 1.891

b) The critical value at 95% confidence interval zα = 1.96 (from z-table)

The calculated z- value < the tabulated value

therefore the null hypothesis accepted

conclusion:-

assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

c) 95% confidence intervals

The confidence intervals are P± 1.96(√PQ/n)

we know that = $p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}$

after calculation we get P = 0.56 and Q =1-P =0.44

Confidence intervals are ( P- 1.96(√PQ/n), P+ 1.96(√PQ/n))

now substitute values , we get

( 0.56- 1.96(√0.56X0.44/700), 0.56+ 1.96(0.56X0.44/700))

on simplification we get (0.523 ,0.596)

Therefore the population proportion (0.56) lies in between the 95% of confidence intervals (0.523 ,0.596)

3 0

3 years ago

(1/3)^x^1/4 • (1/3)^x^2/3

monitta

Wait i’m confused on how this equation is modeled i’m sorry do you have a picture? i wanna be sure i answer this right.

5 0

3 years ago