What is 21 times 15 in standard algorithm

The coordinates for the vertices of a patio are (1,5) (6,5) (6,1) and (1,1). Each grid square has a length of 5 feet. Find the p

nadezda [96]

Answer:

The perimeter of the figure is $90\ ft$

Step-by-step explanation:

Let

$A(1,5),B(6,5),C(6,1),D(1,1)$

using a graphing tool

see the attached figure

The perimeter is equal to

$P=AB+BC+CD+AD$

The figure is a rectangle

so

$AB=DC=(6-1)=5*5=25\ ft$

$AD=BC=(5-1)=4*5=20\ ft$

$P=2(25+20)=90\ ft$

7 0

3 years ago

Read 2 more answers

the cos c of recycle Tin can varies directly as it's weight w in kilograms .if a junk shop pays 60 pesos per kilo of tin cans ho

liraira [26]

Answer:

Step-by-step explanation:

c = 60w

390 = 60w

w = 6.5 kilograms

3 0

3 years ago

Okay can you please help me do this

sertanlavr [38]

Answer:

y=3

Step-by-step explanation:

Multiply 9 from both sides to cancel it out.

-27=-y/9

/9 /9

-3=-y

We can't have a negative variable, so we divide -1 from both sides.

-3=-y

/-1 /-1\

3=y

-hope it helps

7 0

2 years ago

I need help someone PLZZZ HELP ME

Tems11 [23]

Answer:iiikkkkkkkhuf76v76r 7

Step-by-step explanation:12345678910

5 0

3 years ago

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it

brilliants [131]

Answer:

(a) 11.25 and 1.68

(b) 0.1651

(c) 0.3903

(d) 0.6865

Step-by-step explanation:

We are given that GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 75% of the cases and suppose the 15 cases reported today are representative of all complaints.

This situation can be represented through Binomial distribution as;

$P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....$

where, n = number of trials (samples) taken = 15

r = number of success

p = probability of success which in our question is % of cases in

which customer problems are resolved on the same day, i.e.;75%

So, here X ~ $Binom(n=15,p=0.75)$

(a) Expected number of problems to be resolved today = E(X)

E(X) = $\mu$ = n * p = 15 * 0.75 = 11.25

Standard deviation = $\sigma$ = $\sqrt{n*p*(1-p)}$ = $\sqrt{15*0.75*(1-0.75)}$ = 1.68

(b) Probability that 10 of the problems can be resolved today = P(X = 10)

P(X = 10) = $\binom{15}{10}0.75^{10}(1-0.75)^{15-10}$

= $3003*0.75^{10} *0.25^{5}$ = 0.1651

(c) Probability that 10 or 11 of the problems can be resolved today is given by = P(X = 10) + P(X = 11)

= $\binom{15}{10}0.75^{10}(1-0.75)^{15-10}+\binom{15}{11}0.75^{11}(1-0.75)^{15-11}$

= $3003*0.75^{10} *0.25^{5} + 1365*0.75^{11} *0.25^{4}$ = 0.3903

(d) Probability that more than 10 of the problems can be resolved today is

given by = P(X > 10)

P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

= $\binom{15}{11}0.75^{11}(1-0.75)^{15-11}+\binom{15}{12}0.75^{12}(1-0.75)^{15-12} + \binom{15}{13}0.75^{13}(1-0.75)^{15-13}+\binom{15}{14}0.75^{14}(1-0.75)^{15-14} + \binom{15}{15}0.75^{15}(1-0.75)^{15-15}$

= $1365*0.75^{11} *0.25^{4} + 455*0.75^{12} *0.25^{3}+105*0.75^{13} *0.25^{2} + 15*0.75^{14} *0.25^{1}+1*0.75^{15} *0.25^{0}$

= 0.6865

3 0

3 years ago