15 points. Correct answer and you get brainliest answer, 5 stars, and a thanks!!

There are 52 birds and acts in the store. For the same animals Kia counts 138 legs. How many birds and cats are in the pet store

KiRa [710]

Let's represent birds by the letter x and cats by the letter y. Since birds have 2 legs, and cats have 4, we can express this as:
$138 = 2x + 4y$
$52 = x+y$

We can solve by substitution. Let's solve for x in the second equation:
$x = 52-y$

Replace every x in the first equation by that:
$138 = 2(52-y)+4y$
$138 = 104-2y+4y$
$138 = 104 + 2y$
$34 = 2y$
$y = 17$

Now, plug that in for y in second equation and solve for x:
$52 = x + 17$
$x = 35$

Since x was the number of birds and y was the number of cats in the store, there are 35 birds and 17 cats in the store.

4 0

4 years ago

WILL AWARD BRAINLIEST

vovangra [49]

Answer:

hope it helps uh.......

7 0

4 years ago

for the data values below construct a 95 confidence interval if the sample mean is known to be 12898 and the standard deviation

Volgvan

Answer:

A 95% confidence interval for the population mean is [3315.13, 22480.87] .

Step-by-step explanation:

We are given that for quality control purposes, we collect a sample of 200 items and find 24 defective items.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample proportion of defective items = 12,898

s = sample standard deviation = 7,719

n = sample size = 5

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.776 < $t_4$ < 2.776) = 0.95 {As the critical value of t at 4 degrees of

freedom are -2.776 & 2.776 with P = 2.5%}

P(-2.776 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.776) = 0.95

P( $-2.776 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.776 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.776 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.776 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.776 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.776 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $12,898-2.776 \times {\frac{7,719}{\sqrt{5} } }$ , $12,898+2.776 \times {\frac{7,719}{\sqrt{5} } }$ ]

= [3315.13, 22480.87]

Therefore, a 95% confidence interval for the population mean is [3315.13, 22480.87] .

5 0

4 years ago

A cylindrical cake with a height of 5 inches has a 30 degree slice removed. If the volume of the removed slice is 40 inches what

sesenic [268]

Answer: $5.52\ in.$

Step-by-step explanation:

Given

Height of cylindrical cake is $h=5\ in.$

$30^{\circ}$ slice is removed

Volume of the slice is $V=40\ in.^3$

Area of slice is

$\Rightarrow \dfrac{30^{\circ}}{360^{\circ}}\times \pi r^2\\\\\Rightarrow \dfrac{\pi r^2}{12}$

Volume of slice will be multiplication of area and height

$\Rightarrow V=A\times h\\\\\Rightarrow 40=\dfrac{\pi r^2}{12}\times 5\\\\\Rightarrow \pi r^2=\dfrac{40\times 12}{5}\\\\\Rightarrow r^2=\dfrac{40\times 12}{3.142\times 5}\\\\\Rightarrow r^2=30.55\\\Rightarrow r=5.52\ in.$