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Doss [256]
3 years ago
11

11.(A-CED.A.1 ) Write and solve the equation

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
7 0

Answer:

Ahmed runs 6 miles each day

Step-by-step explanation:

∵ Ahmad runs m miles each day

∵  Ali runs twice as many miles as Ahmad

→ That means Ali runs 2 × m miles each day

∴ Ali runs 2m miles each day

∵ Altogether, they run 18 miles each day

→ Add m and 2m, then equate the sum by 18

∴ m + 2m = 18

∴ 3m = 18

→ Divide both sides by 3 to find m

∵ \frac{3m}{3}=\frac{18}{3}

∴ m = 6

∵ m represents the number of miles that Ahmed runs each day

∴ Ahmed runs 6 miles each day

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C it’s I think lol!!!!!^^*
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The sum of two number is 27. One number is 3more than the other number. Write and solve a system of equations to find the two nu
Sladkaya [172]

Answer:

One number is 15 and 3 less than that is 12 so that is your other number

3 0
3 years ago
Data collected over a long period of time show that the length of time x to complete a particular college entrance test is norma
Finger [1]

Using the z-table, the probability that a student taking this test will finish in 100 minutes or less is 0.0824 or 8.24%.

For a normally distributed set of data, given the mean and standard deviation, the probability can be determined by solving the z-score and using the z-table.

First, solve for the z-score using the formula below.

z-score = (x – μ) / σ

where x = individual data value = 100

μ = mean = 125

σ = standard deviation = 18

z-score = (100 – 125) / 18

z-score = (-25) / 18

z-score = -1.39

Find the probability that corresponds to the z-score in the z-table. (see attached images)

-1.39 - (-1.3) : -1.4 - (-1.3) = x - 0.0968 : 0.0808 - 0.0968

-0.09 : -0.1 = x - 0.0968 : -0.016

x - 0.0968 = -0.09(-0.016)/-0.1

x = -0.0144 + 0.0968

x = 0.0824

x = 0.0824

Hence, the probability that a student taking this test will finish in 100 minutes or less is 0.0824 or 8.24%.

Learn more about probability here: brainly.com/question/26822684

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7 0
1 year ago
Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams
hram777 [196]

Answer:

X(16)=25.71grams

Step-by-step explanation:

let X(t) denote grams of C formed in  t mins.

For X grams of C we have:

\frac{2}{3}Xg of A and \frac{1}{3}Xg of B

Amounts of A,B remaining at any given time is expressed as:

40-\frac{2}{3}Xg of A and  50-\frac{1}{3}Xg  of B

Rate at which C is formed satisfies:

\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt  \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}

Apply the initial condition,X(0)=0 ,to the expression above

0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}

Now at X(8)=15:

15=90-\frac{90}{90\times 8k+1}  \ ->75=\frac{90}{720k+1}\\k=0.0002778

Substitute  in X(t) to get

X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71

5 0
3 years ago
"The manager for State Bank and Trust has recently examined the credit card account balances for the customers of her bank and f
yuradex [85]

Answer:

a

   P(X = 4 ) = 0.1876

b

   P(X \le 4)  =  0.8358

Step-by-step explanation:

From the question we are told that

The proportion that has outstanding balance is p = 0.20

The sample size is n = 15

Given that the properties of the binomial distribution apply, for a randomly selected number(X) of credit card

X \  \ ~ Bin (n , p )

Generally the probability of finding 4 customers in a sample of 15 who have "maxed out" their credit cards is mathematically represented as

P(X = 4 ) =  ^nC_4 * p^4 * (1 - p)^{n-4}

=> P(X = 4 ) =  ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}

Here C stand for combination

=> P(X = 4 ) = 0.1876

Generally the probability that 4 or fewer customers in the sample will have balances at the limit of the credit card is mathematically represented as

P(X \le 4) =  [ ^{15}C_0 * (0.20)^0 * (1 - 0.20)^{15-0}]+[ ^{15}C_1 * (0.20)^1 * (1 - 0.20)^{15-1}]+\cdots+[ ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}]

=>   P(X \le 4)  =  0.8358

4 0
3 years ago
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