Assume that 74% of randomly selected adults have a credit card. Assume that a group of five adults is randomly selected. If the group of five adults includes exactly 1 with a credit card, is that value of 1 significantly low

Yes, value of adults having exactly 1 credit card is significantly low.

Step-by-step explanation:

We are given that 74% of randomly selected adults have a credit card. Assume that a group of five adults is randomly selected.

And we have to check that if the group of five adults includes exactly 1 with a credit card, is that value of 1 significantly low or not.

The above situation can be represented through Binomial distribution;

where, n = number of trials (samples) taken = 5 adults

            r = number of success = exactly 1

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

                 having a credit card, i.e; 74%

LET X = Number of adults having a credit card

So, it means X ~

Now, Probability that the group of five adults includes exactly 1 with a credit card is given by = P(X = 1)

                 P(X = 1)  =

                               =

                               = 0.0169

So, the probability that in a group of five adults, exactly 1 have a credit card is 0.0169 or 1.69%.

Now, for any probability value to be significantly low it must be less than 5% as it is considered very unusual or may be called it that the probability of happening of that event is very rare or low.

Since, here our probability is way less than 5% i.e. 1.69%. So we can conclude that the value of adults having exactly 1 credit card in a group of 5 is significantly very low.

62.5 paisa

step-by-step explanation:

4000*70 = 280000

280000-(2000*90) = 100000

4000-2400 = 1600

100000/1600 = 62.5

( give brainliest)

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47