Need help on this one

Bulan rows on a crew team. Her team rows their boat at a split (rate) of 2 min/500 m.

notka56 [123]

Answer:

250

Step-by-step explanation:

If Bulan's team rows their boat at a rate of

2

minutes per

500

meters, they row at a rate of

1

minute per

250

meters. We know this because

1

minute is

1

2

of

2

minutes, and in this time, they will have to have rowed

1

2

the distance they would row in

2

minutes (

500

m).

1

2

of

500

is

250

.

So, we now have the rate in min/m.

If, every

1

minute, Bulan's team rows

250

meters, this means that every

250

meters, they have rowed for

1

minute.

Bulan's team's rowing rate in m/min is

250

m/

1

min.

8 0

3 years ago

Read 2 more answers

Jenny thinks that the expression below is equal to x²-4. If you agree, show that she is correct. If you disagree, show that she

sergiy2304 [10]

Answer:

Jenny is wrong the correct answer is x²+ 4 - 4 x.

Step-by-step explanation:

Given that

(x-2)³∕x-2

(x-2)³∕(x-2) =(x-2) (x-2)²∕(x-2)

(x-2)³∕(x-2) = (x-2)²

As we know that

(a-b)² = a²+b² - 2 ab

So

(x-2)² = x²+ 2² - 2 ˣ 2 ˣ x = x²+ 4 - 4 x

(x-2)² = x²+ 4 - 4 x

It means that

(x-2)³∕(x-2) = (x-2)² = x²+ 4 - 4 x

So Jenny is wrong the correct answer is x²+ 4 - 4 x.

8 0

3 years ago

In the circle below, DB = 22 cm, and m<DBC = 60°. Find BC. Ignore my handwriting.

Karo-lina-s [1.5K]

Answer:

$BC=11\ cm$

Step-by-step explanation:

step 1

Find the measure of the arc DC

we know that

The inscribed angle measures half of the arc comprising

$m\angle DBC=\frac{1}{2}[arc\ DC]$

substitute the values

$60\°=\frac{1}{2}[arc\ DC]$

$120\°=arc\ DC$

$arc\ DC=120\°$

step 2

Find the measure of arc BC

we know that

$arc\ DC+arc\ BC=180\°$ ----> because the diameter BD divide the circle into two equal parts

$120\°+arc\ BC=180\°$

$arc\ BC=180\°-120\°=60\°$

step 3

Find the measure of angle BDC

we know that

The inscribed angle measures half of the arc comprising

$m\angle BDC=\frac{1}{2}[arc\ BC]$

substitute the values

$m\angle BDC=\frac{1}{2}[60\°]$

$m\angle BDC=30\°$

therefore

The triangle DBC is a right triangle ---> 60°-30°-90°

step 4

Find the measure of BC

we know that

In the right triangle DBC

$sin(\angle BDC)=BC/BD$

$BC=(BD)sin(\angle BDC)$

substitute the values

$BC=(22)sin(30\°)=11\ cm$

6 0

3 years ago

Th e library fi ne for an overdue DVD is $2 for the fi rst day plus $0.25 for each additional day. How many days overdue is a DV

Mumz [18]

The DVD is 6 days overdue

6 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