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lozanna [386]
3 years ago
11

What is 61% of 524? Thanks

Mathematics
2 answers:
seropon [69]3 years ago
4 0

Answer:

319.64

Step-by-step explanation:

524/100= 5.24

5.24x61= 319.64

Vanyuwa [196]3 years ago
3 0
319.64 i think hope this helps
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Round 187.890 to the nearest 10 000
stellarik [79]

Answer:

190,000

190,000

190,000

Step-by-step explanation:

190,000

190,000

6 0
2 years ago
32^(x+4) = 64^(2x-1)
Rudik [331]

Answer:

Vamos a resolver la ecuación paso a paso.

32x+4=642x−1

Step-by-step explanation:

Resolver el exponente.

32x+4=642x−1

log(32x+4)=log(642x−1)(Sacar el logaritmo de ambos lados.)

(x+4)*(log(32))=(2x−1)*(log(64))

x+4=(

log(64)

log(32)

)*(2x−1)

x+4=1.2*(2x−1)

x+4=2.4x−1.2(Simplificar ambos lados de la ecuación)

x+4−2.4x=2.4x−1.2−2.4x(Restar 2.4x a ambos lados)

−1.4x+4=−1.2

−1.4x+4−4=−1.2−4(Restar 4 a ambos lados)

−1.4x=−5.2

−1.4x

−1.4

=

−5.2

−1.4

(Dividir ambos lados por -1.4)

x=3.714286

Solución:

x=3.714286

7 0
2 years ago
- How many females represented in the Venn diagram below are right-handed?
Wewaii [24]

Answer:

You did not include the Venn diagram

Step-by-step explanation:

3 0
2 years ago
g A population is infected with a certain infectious disease. It is known that 95% of the population has not contracted the dise
trasher [3.6K]

Answer:

There is approximately 17% chance of a person not having a disease if he or she has tested positive.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = a person has contracted the disease.

+ = a person tests positive

- = a person tests negative

The information provided is:

P(D^{c})=0.95\\P(+|D) = 0.98\\P(+|D^{c})=0.01

Compute the missing probabilities as follows:

P(D) = 1- P(D^{c})=1-0.95=0.05\\\\P(-|D)=1-P(+|D)=1-0.98=0.02\\\\P(-|D^{c})=1-P(+|D^{c})=1-0.01=0.99

The Bayes' theorem states that the conditional probability of an event, say <em>A</em> provided that another event <em>B</em> has already occurred is:

P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^{c})P(A^{c})}

Compute the probability that a random selected person does not have the infection if he or she has tested positive as follows:

P(D^{c}|+)=\frac{P(+|D^{c})P(D^{c})}{P(+|D^{c})P(D^{c})+P(+|D)P(D)}

              =\frac{(0.01\times 0.95)}{(0.01\times 0.95)+(0.98\times 0.05)}\\\\=\frac{0.0095}{0.0095+0.0475}\\\\=0.1666667\\\\\approx 0.1667

So, there is approximately 17% chance of a person not having a disease if he or she has tested positive.

As the false negative rate of the test is 1%, this probability is not unusual considering the huge number of test done.

7 0
3 years ago
A business journal investigation of the performance and timing of corporate acquisitions discovered that in a random sample of 2
weeeeeb [17]

Answer:

z = 1.960

Step-by-step explanation:

The sample proportion is:

p = 715 / 2684 = 0.2664

The standard error is:

σ = √(pq/n)

σ = √(0.266 × 0.734 / 2684)

σ = 0.0085

For α = 0.05, the confidence level is 95%.  The z-statistic at 95% confidence is 1.960.

The margin of error is 1.960 × 0.0085 = 0.0167.

The confidence interval is 0.2664 ± 0.0167 = (0.2497, 0.2831).

The upper limit is 28.3%, so the journal can conclude with 95% confidence that the true percentage is less than 29%.

4 0
3 years ago
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