True or false? The first distribution shown below has a smaller mean than does the second.

QRST is a square. PQ = 2√ RU = 4<br> What is the length of SU? Round to the nearest tenth.

777dan777 [17]

Answer:

$SU=3.5\ units$

Step-by-step explanation:

step 1

In the isosceles right triangle PQT

PQ=PT ----> because is an isosceles triangle

Applying the Pythagoras Theorem

we have

$PQ=PT =\sqrt{2}\ units$

$QT^{2}=PQ^{2} +PT^{2}$

substitute the values

$QT^{2}=\sqrt{2}^{2} +\sqrt{2}^{2}$

$QT^{2}=4$

$QT=2\ units$

step 2

In the square QRST

$RS=QT=2\ units$

step 3

In the right triangle RSU

Applying the Pythagoras Theorem

$RU^{2}=RS^{2} +SU^{2}$

we have

$RU=4\ units$

$RS=2\ units$

substitute the values and solve for SU

$4^{2}=2^{2} +SU^{2}$

$SU^{2}=4^{2}-2^{2}$

$SU^{2}=12$

$SU=2\sqrt{3}\ units$

$SU=3.5\ units$

8 0

3 years ago

Factor 15abc − 9ab − 3bcd completely

goldenfox [79]

I believe the correct answer is C

3 0

3 years ago

Read 2 more answers

A grocery store offers a 15% discount on your entire purchase when you spend more than $60. You have $85 to spend on groceries a

Serga [27]

0.85(a + 60) ≤ 85

0.85a + 51 ≤ 85

-51 on both sides

0.85a ≤ 34

/0.85 on both sides

a ≤ 40

Answer:

a ≤ 40

5 0

3 years ago

In homer run derby competition there were 32 home rinse hit out of 124 hits approximately what percent of hits were home runs

RoseWind [281]

Answer:

26% of hits were home runs.

Step-by-step explanation:

In homer run derby competition there were 32 home rinse hit out of 124 hits approximately.

We need to find what percentage of hits were home runs.

$\text{Formula for percentage }=\frac{\text{Number of hit by home run}}{\text{Total number of hits}}\times 100$

Number of home runs hit = 32

Total number of hit = 124

Now we calculate the percentage

$\%=\frac{32}{124}\times 100\approx 26$

Thus, 26% of hits were home runs.

3 0

3 years ago

Una empresa exportadora de frutas vende dos variedades de frambuesas: estándar y de lujo. Una caja de freses estándar se vende e

aksik [14]

Answer:

83.33 cajas estándar de frambuesas y 51.67 cajas de lujo de frambuesas.

Step-by-step explanation:

Representemos el número de cajas como

A = caja estándar de frambuesas

B = caja de lujo de frambuesas

Caja estándar de frambuesas = $ 7 Caja de lujo de frambuesas = 10.

A + B = 135 ......... Ecuación 1

B = 135 - A

7A + 10B = 1100 ........... Ecuación 2

Sustituir

135 - A para B en la ecuación 2

7A + 10 (135 - A) = 1100

7A + 1350 -10A = 1100

7A - 10A = 1100-1350

-3A = - 250

A = 250/3

A = 83.33 cajas

Sustituye 83.33 por A en la ecuación 1

A + B = 135

83,33 + B = 135

B = 135 - 83.33 = 51.67 cajas

Por lo tanto, vendió,

83.33 cajas estándar de frambuesas y 51.67 cajas de lujo de frambuesas.

6 0

3 years ago