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Softa [21]
1 year ago
12

Diego has $500 in his savings account. each month, he takes out $20 to pay for karate class.

Mathematics
1 answer:
lyudmila [28]1 year ago
7 0

Answer:

25m?

Step-by-step explanation:

I mean I'm not sure what the question is asking for but after 25 months he won't have any money left lol hope that helped ♡

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Given: PRST square
Zigmanuir [339]

Answer:

7a²/16

Step-by-step explanation:

Area of the triangle PTS

½ × a × a

a²/2

Length of PS:

sqrt(a² + a²)

asqrt(2)

Length of MS:

¼asqrt(2)

Triangles MCS and TPS are similar

With sides in the ratio:

¼asqrt(2) : a

sqrt(2)/4 : 1

Area of triangle SMC:

A/(a²/2) = [(sqrt(2)/4)]²

2A/a² = 1/8

A = a²/16

Area of PTMC

= a²/2 - a²/16

= 7a²/16

Step-by-step explanation:

4 0
1 year ago
Read 2 more answers
Of the 500 sample households in the previous exercise, 7 had three or more large-screen TVs. (a) The percentage of households in
likoan [24]

Answer:

a) The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

\hat p=\frac{7}{500}=0.014

And that represent the 1.4%.

b) And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

Step-by-step explanation:

Data given and notation  

n=500 represent the random sample taken    

X=7 represent the households with three or more large-screen TVs

\hat p=\frac{7}{500}=0.014 estimated proportion of households with three or more large-screen TVs

\alpha=0.05 represent the significance level (no given, but is assumed)    

z would represent the statistic (variable of interest)    

p= population proportion of households with three or more large-screen TVs

Part a

The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

\hat p=\frac{7}{500}=0.014

And that represent the 1.4%.

Part b

Yes is possible. We hav that np>10 and n(1-p)>10 so we have the assumption of normality to find the interval.

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

And replacing into the confidence interval formula we got:

0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370

0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243

And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

4 0
1 year ago
A card is drawn from a deck of cards numbered one through twenty. The card is NOT replaced and another card is drawn. Find the p
Bess [88]

Answer:

B)  P(two prime numbers are drawn in a row) = 14/95

Step-by-step explanation:

Total cards in the deck  = 20

Total prime numbered cards in deck  = { 2, 3, 5, 7, 11, 13, 17, 19}  = 8 cards

So, Probability of picking two prime cards from deck (without replacing)

= Probability of picking first prime card x Probability of picking second prime card

P( picking first prime card ) = \frac{\textrm{Total prime cards in the deck}}{\textrm{Total available cards}}

= \frac{8  }{20}  = \frac{2}{5}

P( picking second prime card )  = \frac{\textrm{Total prime cards in the deck}}{\textrm{Total available cards}}

=\frac{7}{19}

Hence, the total probability =\frac{2}{5}  \times \frac{7}{19} = \frac{14}{95}

or, B)  P(two prime numbers are drawn in a row) = 14/95

4 0
1 year ago
The table gives the probabilities that orphaned pets in animal shelters in six cities are one of the types listed.
vagabundo [1.1K]

Answer:

the ablity to choose one is the right now

Step-by-step explanation:

3 0
1 year ago
If Machine A makes a yo-yo every five minutes and Machine B takes ten minutes to make a yo-yo, how many hours would it take them
Murrr4er [49]
If every 5 mins, A makes 1 yo-yo every 10 mins, B makes 1 yo-yo then every 10 mins, both machines produce 3 yo-yos every 10 mins (2 from machine A and 1 from machine B) Therefore, for 20 yo-yos, both machines would take 70 minutes( 1 hour and 10 mins). After 70 minutes, 21 yo-yos would be produced.
7 0
1 year ago
Read 2 more answers
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